# Toronto Math Forum

## APM346-2012 => APM346 Math => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2012, 06:29:34 PM

Title: TT1 = Problem 4
Post by: Victor Ivrii on October 16, 2012, 06:29:34 PM
Consider the initial value problem for the diffusion equation on the line:
\begin{equation*}
\left\{\begin{aligned}
&u_{t} = k  u_{xx},\qquad&&x \in \mathbb{R},\\
&u (x,0) = f(x), \qquad&&x \in \mathbb{R}.
\end{aligned}\right.
\end{equation*}
• (a) Assuming $f$ is smooth and vanishes when $|x|>10$, give a formula for the solution $u(x,t)$.
• (b) Using the formula from part (a), prove that
\begin{equation*}
\lim_{t \searrow 0} u(x,t) = f(x).
\end{equation*}
Title: Re: TT1 = Problem 4
Post by: Ian Kivlichan on October 16, 2012, 07:03:44 PM
Solution to 4.a) attached!

Re-written more nicely. Original at http://i.imgur.com/M2yhk.jpg
Title: Re: TT1 = Problem 4
Post by: Djirar on October 16, 2012, 08:15:40 PM
part b.
Title: Re: TT1 = Problem 4
Post by: Jinchao Lin on October 16, 2012, 09:11:35 PM
From part (a) we have
$u(x,t) = \frac{1}{\sqrt{4\pi kt}} \int_{-10}^{10} e^{ \frac{-(x-y)^2}{4kt} } f(y) dy$

Rewrite the formula as:
$u(x,t) = \frac{1}{\sqrt{2\pi } \sqrt{2kt}} \int_{-10}^{10} e^{ \frac{-(x-y)^2}{2 \sqrt{2kt}^2 } } f(y) dy$

We see this is a normal density of random variable x centered at y with standard error $\sqrt{2kt}$ .  When $t \to 0$, the standard error approach to 0 as well. The random variable x approaches to a deterministic form. So $lim_{t \to 0} u(x,t) = f(y)$.

I think the argument can be written more accurate if we take the functional form of f(x) in the whole real line. But then we are not directly prove the result based on the formula of u(x,t) we get from part (a).
Title: Re: TT1 = Problem 4
Post by: Victor Ivrii on October 16, 2012, 09:57:48 PM
Both solutions are correct.

Remark: $x$ is not a random variable. This function may represent density of the distribution of the random variable $\xi$.
Title: Re: TT1 = Problem 4
Post by: Bowei Xiao on October 17, 2012, 11:54:35 AM
Is it ok to use he fact that u is actually continious and so the limit is actually the initially boundary condition?
Title: Re: TT1 = Problem 4
Post by: Danny Dinh on October 17, 2012, 04:41:43 PM
Is it ok to use he fact that u is actually continious and so the limit is actually the initially boundary condition?

The question says to use the formula from part (a).
Title: Re: TT1 = Problem 4
Post by: Bowei Xiao on October 18, 2012, 12:23:17 AM
Is it ok to use he fact that u is actually continious and so the limit is actually the initially boundary condition?

The question says to use the formula from part (a).
the formula from a can lead to continuity as well.. I guess...omg..hope they give me some partial marks!