# Toronto Math Forum

## APM346-2018S => APM346--Tests => Term Test 1 => Topic started by: Victor Ivrii on February 15, 2018, 07:08:45 PM

Title: P5
Post by: Victor Ivrii on February 15, 2018, 07:08:45 PM
$\newcommand{erf}{\operatorname{erf}}$
Find the solution $u(x,t)$ to
\begin{align}
&u_t=u_{xx} && -\infty<x<\infty, \ t>0,
\tag{1}\\[2pt]
&u|_{t=0}=e^{-|x|}
\tag{2}\\
&\max |u|<\infty.
\tag{3}
\end{align}
Calculate the integral.

Hint: For $u_t=ku_{xx}$ use

G(x,y,t)=\frac{1}{\sqrt{4\pi kt}}\exp (- (x-y)^2/4kt).
\tag{4}

To calculate integral make change of variables and use $\erf(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-z^2}\,dz$.
Title: Re: P5
Post by: Jingxuan Zhang on February 16, 2018, 02:50:23 PM
A direct computation yields:
\begin{align*}

u&=\frac{1}{\sqrt{4t\pi}}\int_{-\infty}^{\infty} \exp{-\frac{(y-x)^2}{4t}-|y|}\,dx\\
&=\frac{1}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y+x)^2}{4t}-y)\,dx +\frac{1}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y-x)^2}{4t}-y)\,dx\\
&=\frac{\exp(x+t)}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y+x+2t)^2}{4t})\,dx + \frac{\exp(-x+t)}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y-x+2t)^2}{4t})\,dx\\
&=\frac{\exp(x+t)}{\sqrt{\pi}}\int_{\frac{x+2t}{\sqrt{4t}}}^{\infty} e^{-z^2} \,dz+ \frac{\exp(-x+t)}{\sqrt{\pi}}\int_{\frac{-x+2t}{\sqrt{4t}}}^{\infty} e^{-z^2} \,dz\\
&=\frac{\exp(x+t)}{2}(1-\text{erf}(\frac{x+2t}{\sqrt{4t}})) + \frac{\exp(-x+t)}{2}(1-\text{erf}(\frac{-x+2t}{\sqrt{4t}}))
\end{align*}
Title: Re: P5
Post by: Tristan Fraser on February 16, 2018, 03:15:35 PM
This solution can be split into two parts: for $y> 0$ and $y< 0$, yielding

$$e^{-x}$$
$$e^{x}$$ respectively.

So u is split over those two regions. For the first one, we have

$$u(x,t) = \frac{1}{\sqrt{4\pi t}} \int_0^{\infty} \exp(\frac{-(x-y)^2) - 4yt}{4t}) dy$$

This calls for completing the square, then integrating as usual:

$$u(x,t) = \frac{1}{\sqrt{4\pi t}} \int_0^{\infty} \exp(\frac{-(x-y)^2) - 4yt}{4t})$$

$$u(x,t) = \frac{1}{\sqrt{4\pi t}} \int_0^{\infty} \exp(\frac{-(y-(x-2t)^2)}{4t} exp(t-x) dy$$

$$u(x,t) = \frac{exp(t-x)}{\sqrt{4\pi t}} \int_0^{\infty} \exp(\frac{-(y-(x-2t)^2)}{4t}$$

Now use the error function $\erf(z) = \frac{2}{\sqrt{\pi}} \int_{0}^{z} e^{-z^2} dz$ to answer:
let $z = (y -(x-2t)) / \sqrt{4t}$ and $dz = \frac{dy}{\sqrt{4t}}$ giving us the integral:

$$u(x,t) = \frac{\sqrt{4t} exp(t-x)}{\sqrt{4\pi t}}\int_{\frac{2t-x}{\sqrt{4t}}}^{\infty} e^{-z^2}dz$$

$$u(x,t) = \frac{exp(t-x)}{2} erf(\infty) - \frac{\exp(t-x)}{2}\erf(\frac{2t-x}{\sqrt{4t}})$$

Repeating these steps for the other region with $\phi(x) = e^x$ gives us:

$$u(x,t) = -\frac{exp(t+x)}{2} \erf(-\infty) + \frac{\exp(t+x)}{2}erf(\frac{-2t-x}{\sqrt{4t}})$$
\
Note that error function at $\pm \infty = \pm 1$

Adding these together should give the solutions that Jingxuan posted.
Title: Re: P5
Post by: Victor Ivrii on February 22, 2018, 07:01:40 AM
Tristan
$u$ is not split , but the integral, expressinfg it is