Author Topic: TT1-P5  (Read 1232 times)

Victor Ivrii

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TT1-P5
« on: October 21, 2015, 08:52:22 PM »
$\newcommand{\erf}{\operatorname{erf}}$
Find  solution $u(x,t)$ to
\begin{align}
&u_t=u_{xx} && -\infty<x<\infty, \ t>0,\label{eq-5-1}\\[4pt]
&u|_{t=0}=\left\{\begin{aligned}
&x\qquad && |x|<1,\\[2pt]
&0      && |x|>1
\end{aligned}\right.,\\[4pt]
&\max |u|<\infty.
\end{align}
Calculate integral.


Hint:
For $u_t=ku_{xx}$ use $\displaystyle{G(x,t)=\frac{1}{\sqrt{4\pi kt}}\exp (- (x-y)^2/4kt)}$.
To calculate integral make change of variables and use $\displaystyle{\erf(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-z^2}\,dz}$.
« Last Edit: October 21, 2015, 08:53:53 PM by Victor Ivrii »

Yeming Wen

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Re: TT1-P5
« Reply #1 on: October 21, 2015, 11:28:16 PM »
Notice we only need to integrate $$\frac{1}{\sqrt{4\pi kt}}\int^1_{-1} ye^{-\frac{(x-y)^2}{4kt}}dy$$
Then we change of variable $z=\frac{y-x}{\sqrt{4kt}}$.
So the integral becomes
\begin{equation*}
\begin{split}
\frac{1}{\sqrt{4\pi kt}}\int^a_b\sqrt{4kt}(x+\sqrt{4t}z)e^{-z^2}dz &= \frac{1}{\sqrt{\pi}}\int^a_b (x+2\sqrt{t}z)e^{-z^2}dz \\
&= \frac{x}{\sqrt{\pi}}\int^a_b e^{-z^2}dz+\frac{1}{\sqrt{\pi}}\int^a_b 2\sqrt{t}ze^{-z^2}dz\\
&= \frac{x}{2}[E(a)-E(b)] +\frac{2\sqrt{t}}{\sqrt{\pi}}(-\frac{1}{2}e^{-z^2}) \rvert^a_b\\
&= \frac{x}{2}[E(a)-E(b)] + \sqrt{\frac{t}{\pi}}[e^{-\frac{(1+x)^2}{4t}}-e^{-\frac{(1-x)^2}{4t}}]
\end{split}
\end{equation*}
where $$a=\frac{1-x}{2\sqrt{t}}$$ and $$b=\frac{-1-x}{2\sqrt{t}}$$ and $E$ is the error function.
« Last Edit: October 22, 2015, 12:14:31 PM by Yeming Wen »