Section 2.1 #1(a)

Mark

Solve the Cauchy problem
$$
\begin{align}
u_t &= u_{xx}, \text{ over } x\in{\mathbb R}, t>0 \\
u(x,0) &= \left\{
\begin{array}{ll}
1 & \text{if } |x|\leq 1 \\
0 & \text{if } |x|>1.
\end{array}
\right.
\end{align}
$$
Write your answer down in terms of an integral and attempt to express the integral in terms of the error function. Sketch the initial condition, the steady state, and the evolution towards the steady state.

sfrye

So I believe that this will satisfy the "answer in terms of an integral" part of the question.
$$u(x,t) = \frac{1}{\sqrt{4\pi t}}\int_{-1}^{1} e^{-(x-y)^2/(4t)} dy$$

Here are some sketches of the progression.

cromer

To ensure LaTeX will raise the entire expression into the superscript, you need to enclose it in curly brackets. For example, with the code

$$ e^{ -(x-y)^2 / 4t } .$$

you get

$$ e^{-(x-y)^2/4t} .$$

There are some handy guides which are suited for when you need to find a specific piece of LaTeX formatting. This Wikibook has most of the stuff you might want to do on the fly, and you can also almost always find something helpful just by Googling "(thing you want to do) latex".

Your answers look spot on to me.

Mark

@sfrye I combined your posts into one, since they really address the same question, and corrected your formatting. Also, the integral should be $dy$, rather than $dx$.

sfrye

Thanks Mark. the question also asks for the integral in terms of the error function. But the integral already looks very similar to the error function. Am I missing a step to get from one to the other?

sfrye

So...
Using the u substitution we reviewed in class today,
$$u=\frac{x-y}{2\sqrt{t}}$$
$$y=-1$$ becomes $$u=\frac{x+1}{2\sqrt{t}}$$
and $$y=1$$ becomes $$u=\frac{x-1}{2\sqrt{t}}$$
Using these bounds, our function becomes:
$$\frac{1}{2}\left[ \mathrm{erf}\left(\frac{x+1}{2\sqrt{t}} \right) -\mathrm{erf}\left(\frac{x-1}{2 \sqrt{ t } } \right) \right]$$

Mark

Much, much better!