### Author Topic: TT1 = Problem 4  (Read 10724 times)

#### Victor Ivrii ##### TT1 = Problem 4
« 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*}
« Last Edit: October 16, 2012, 06:57:16 PM by Victor Ivrii »

#### Ian Kivlichan

• Sr. Member
•    • Posts: 51
• Karma: 17 ##### Re: TT1 = Problem 4
« Reply #1 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
« Last Edit: October 17, 2012, 01:09:57 AM by Ian Kivlichan »

#### Djirar

• Full Member
•   • Posts: 24
• Karma: 8 ##### Re: TT1 = Problem 4
« Reply #2 on: October 16, 2012, 08:15:40 PM »
part b.

#### Jinchao Lin

• Jr. Member
•  • Posts: 14
• Karma: 6 ##### Re: TT1 = Problem 4
« Reply #3 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).

#### Victor Ivrii ##### Re: TT1 = Problem 4
« Reply #4 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 $\xi$.

#### Bowei Xiao

• Full Member
•   • Posts: 17
• Karma: 2 ##### Re: TT1 = Problem 4
« Reply #5 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?

#### Danny Dinh

• Newbie
• • Posts: 3
• Karma: 1 ##### Re: TT1 = Problem 4
« Reply #6 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).

#### Bowei Xiao

• Full Member
•   • Posts: 17
• Karma: 2 ##### Re: TT1 = Problem 4
« Reply #7 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!