FE-P2

[Victor Ivrii]

$\newcommand{\erf}{\operatorname{erf}}$
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]

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

[Victor Ivrii]

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 $u(x,1)$.
General comments
Typical errors:
* Solving IVP rather than IBVP
* Improper square separation mentioned in (*)
* Forgetting to change the lower limit in $\int_0^\infty \ldots dy$ while changing variable $z= (x-y \pm c t)/\sqrt{2t}$.