### Author Topic: P5 Night  (Read 1045 times)

#### Victor Ivrii

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##### P5 Night
« on: February 15, 2018, 08:27:08 PM »
$\newcommand{\erf}{\operatorname{erf}}$
Find the solution $u(x,t)$ to
\begin{align}
&u_t=u_{xx} && -\infty<x<\infty, \ t>0,\tag{1}\\[2pt]
&u|_{t=0}=\left\{\begin{aligned} &x &&|x|<1,\\ &0 &&|x|\ge 1,\end{aligned}\right.\tag{2}\\
&\max |u|<\infty.\tag{3}
\end{align}
Calculate the integral.

Hint: For $u_t=ku_{xx}$ use

G(x,y,t)=\frac{1}{\sqrt{4\pi kt}}\exp (- (x-y)^2/4kt).
\tag{4}

To calculate integral make change of variables and use $\erf(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-z^2}\,dz$.

#### Jilong Bi

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##### Re: P5 Night
« Reply #1 on: February 15, 2018, 10:03:55 PM »
We know k = 1 ,so by formular:  $$\frac{1}{\sqrt{4{\pi}t}}\int_{-1}^1 e^{-\frac{(x-y)^2}{4t}}ydy$$  Let$$w = {\frac{(y-x)}{\sqrt{4t}}} \Rightarrow y =\sqrt{4t}w+x \Rightarrow dy = \sqrt{4t} dw$$ then the formular becomes  $$\frac{1}{\sqrt{4{\pi}t}}\int_{b}^a e^{-w^2}(\sqrt{4t}w+x) \sqrt{4t} dw \Rightarrow \frac{1}{\sqrt{{\pi}}}\int_{b}^a e^{-w^2}(\sqrt{4t}w+x) dw \Rightarrow \frac{1}{\sqrt{{\pi}}}\int_{b}^a \sqrt{4t}we^{-w^2} dw+ \frac{1}{\sqrt{{\pi}}}\int_{b}^a xe^{-w^2} dw$$  $$\frac{2\sqrt{t}}{\sqrt{{\pi}}}(-\frac{1}{2}e^{-a^2}+\frac{1}{2}e^{-b^2}) +\frac{x}{2}[erf(a)-erf(b)]$$ $$\Rightarrow\frac{\sqrt{t}}{\sqrt{{\pi}}} (\frac{1}{2}e^{-b^2}-\frac{1}{2}e^{-a^2} +\frac{x}{2}[erf(a)-erf(b)]$$ $$a = {\frac{1 -x}{\sqrt{4t}}}$$ $$b ={\frac{-1-x}{\sqrt{4t}}}$$