### Author Topic: FE-P2  (Read 984 times)

#### Victor Ivrii ##### FE-P2
« on: April 11, 2018, 02:34:19 PM »
$\newcommand{\erf}{\operatorname{erf}}$
Solve  IVP for the heat equation
\begin{align}
&2u_t -   u_{xx}=0,\qquad &&0 <x<\infty,\; t>0,\label{2-1}\\[2pt]
&u|_{x=0}=0,\\
&u|_{t=0}= f(x)\label{2-2}
\end{align}
with $f(x)=e^{-x}$.

Solution should be expressed  through $\displaystyle{\erf(z)=\frac{2}{\sqrt{\pi}} \int_0^z e^{-z^2}\,dz}$

#### Andrew Hardy

• Full Member
•   • Posts: 34
• Karma: 10 ##### Re: FE-P2
« Reply #1 on: April 11, 2018, 05:23:54 PM »
Very similar to Term Test 1. Here instead though,  apply even continuation and then because k = 1/2

$$u=\frac{1}{\sqrt{2t\pi}}\int_0^{\infty} \exp(-\frac{(y+x)^2}{2t}-y) \,dx +\frac{1}{\sqrt{t\pi}}\int_0^{\infty} \exp(-\frac{(y-x)^2}{2t}-y)\,dx\\$$
and then completing the square
$$=\frac{\exp(x+t)}{\sqrt{t\pi}}\int_0^{\infty} \exp(-\frac{(y+x+t)^2}{2t})\,dx + \frac{\exp(-x+t)}{\sqrt{t\pi}}\int_0^{\infty} \exp(-\frac{(y-x+t)^2}{2t})\,dx\\$$
and then via change of variables
$$=\frac{\exp(x+t/2)}{\sqrt{\pi}}\int_{\frac{x+t/2}{\sqrt{2t}}}^{\infty} e^{-z^2} \,dz + \frac{\exp(-x+t/2)}{\sqrt{\pi}}\int_{\frac{-x+t}{\sqrt{2t}}}^{\infty} e^{-z^2} \,dz\\$$
and in conclusion
$$=\frac{\exp(x+t/2)}{2}(1-\text{erf}(\frac{x+t}{\sqrt{2t}})) + \frac{\exp(-x+t/2)}{2}(1-\text{erf}(\frac{-x+t}{\sqrt{2t}}))$$

corrected
« Last Edit: April 12, 2018, 03:45:09 PM by Andrew Hardy »

#### Victor Ivrii ##### Re: FE-P2
« Reply #2 on: April 12, 2018, 01:08:09 PM »
This looks familiar:)
Indeed, it looks familiar but in addition to misprints there are errors, leading to the errors in the answer.
Jingxuan, you are the second most prolific poster on this forum, you just made more than Emily, but this was a flood. Deleted.

$$\frac{1}{2t} (x+y )^2 + y \overset{?}{=} \frac{1}{2t} (x+y +{\color{red}{2}}t)^2 - ... \tag{*}$$
Now it is fixed. I sketched $u(x,0)$ and $u(x,1)$.
* Forgetting to change the lower limit in $\int_0^\infty \ldots dy$ while changing variable $z= (x-y \pm c t)/\sqrt{2t}$.