# Toronto Math Forum

## APM346-2016F => APM346--Tests => FE => Topic started by: Victor Ivrii on December 13, 2016, 07:50:03 PM

Title: FE2
Post by: Victor Ivrii on December 13, 2016, 07:50:03 PM
$\newcommand{\erf}{\operatorname{erf}}$
Solve  IVP for the heat equation
\begin{align}
&4u_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 $f(x)=xe^{-x^2}$.

Solution should be expressed  through $\displaystyle{\erf(z)=\frac{2}{\sqrt{\pi}} \int_0^z e^{-z^2}\,dz}$.
Title: Re: FE2
Post by: Victor Ivrii on December 15, 2016, 08:00:35 AM
Wrong solution removed. Several students got correct solutions, including calculations. I hope that  one of them posts (or I will).
Title: Re: FE2
Post by: Luyu CEN on December 15, 2016, 06:36:09 PM
This is 1D heat equation on half-line problem with Dirichlet boundary condition.
\begin{equation}
U(x,t)=\int
_{-\infty}^\infty G(x,y,t) F(y)\,dy.
\end{equation}
where $F(y)$ is the function by taking the odd continuation of $f(y)$ and
\begin{equation}
G(x,y,t)=
\frac{1}{2\sqrt{k\pi t}}e^{-\frac{(x-y)^2}{4kt}}
\end{equation}
Note that $f(y)$ is actually an odd function itself and substitute k with $\frac{1}{4}$, we have
\begin{equation}
U(x,t)=\int
_{-\infty}^\infty G(x,y,t) f(y)\,dy = \int
_{-\infty}^\infty \frac{1}{\sqrt{\pi t}}e^{-\frac{(x-y)^2}{t}} ye^{-y^2}\,dy
\end{equation}
\begin{multline*}
\int
_{-\infty}^\infty \frac{1}{\sqrt{\pi t}}e^{-\frac{(x-y)^2}{t}} ye^{-y^2}\,dy
= \frac{1}{\sqrt{\pi t}}e^{-\frac{x^2}{1+t}}\int
_{-\infty}^\infty ye^{-\frac{(1+t)(y-\frac{x}{1+t})^2}{t}} \,dy
\\=\frac{1}{\sqrt{\pi t}}e^{-\frac{x^2}{1+t}}\int
_{-\infty}^\infty \bigl(\frac{x}{1+t}e^{-\frac{(1+t)(y-\frac{x}{1+t})^2}{t}} + (y-\frac{x}{1+t})e^{-\frac{(1+t)(y-\frac{x}{1+t})^2}{t}}\bigr)\,dy
\\= \frac{1}{\sqrt{\pi t}}\sqrt{\frac{t}{1+t}}\frac{\sqrt{\pi}}{2}\frac{x}{1+t}e^{-\frac{x^2}{1+t}}(erf(\infty)-erf(-\infty)) + \frac{1}{\sqrt{\pi t}}e^{-\frac{x^2}{1+t}}\sqrt{\frac{t}{1+t}}e^{-\frac{(1+t)(y-\frac{x}{1+t})^2}{t}}|_{-\infty}^{\infty} \\= \frac{xe^\frac{-x^2}{1+t}}{\sqrt{1+t}(1+t)}
\end{multline*}
Therefore, the solution is given by
\begin{equation}
u(x,t) = \frac{xe^\frac{-x^2}{1+t}}{\sqrt{1+t}(1+t)}
\end{equation}
Title: Re: FE2
Post by: Victor Ivrii on December 15, 2016, 07:42:07 PM
Indeed, $u=\frac{x}{(t+1)^{\frac{3}{2}}}e^{-\frac{x^2}{(t+1)}}$ and there is a way to get it even almost without calculations. First, as mentioned, since $f(x)=xe^{-x^2}$ is an odd function, it is sufficient to solve Cauchy problem.

We know that $v(x,t)=t^{-\frac{1}{2}}e^{-\frac{x^2}{t}}$ satisfies our equation (\ref{2-1}).
Plugging $t:=t+1$ we conclude that $w(x,t)=(t+1)^{-\frac{1}{2}} e^{-\frac{x^2}{t+1}}$ also satisfies (\ref{2-1}) (since coefficients do not depend on $t$). Further, $w(x,0)=g(x):=e^{-x^2}$.

Next, $-\frac{1}{2}w_x$ also satisfies (\ref{2-1}) (since coefficients do not depend on $x$) and $-\frac{1}{2}w_x(x,0)=-\frac{1}{2}g'(x)$ which is $f(x)$. Bingo! $u=-\frac{1}{2}w_x=\frac{x}{(t+1)^{\frac{3}{2}}}e^{-\frac{x^2}{(t+1)}}$.