### Author Topic: FE-2  (Read 2854 times)

#### Victor Ivrii

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##### FE-2
« on: December 18, 2015, 07:41:43 PM »
$\newcommand{\erf}{\operatorname{erf}}$
Solve  IVP for the heat equation
\begin{align}
&u_t - u_{xx}=0,\qquad &&-\infty <x<\infty,\; t>0,\label{2-1}\\[2pt]
&u|_{t=0}= f(x)\label{2-2}
\end{align}
with $f(x)=|x|$.

Solution should be expressed  through $\displaystyle{\erf(z)=\frac{2}{\sqrt{\pi}} \int_0^z e^{-z^2}\,dz}$.
« Last Edit: December 20, 2015, 06:32:14 AM by Victor Ivrii »

#### Xi Yue Wang

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##### Re: FE-2
« Reply #1 on: December 18, 2015, 11:02:01 PM »
we have $$u(x,t) = \frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}[e^\frac{-(x-y)^2}{4t} + e^\frac{-(x+y)^2}{4t}]y dy$$
To solve the first part, we get $$\frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x-y)^2}{4t}y dy = \frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x-y)^2}{4t}(y-x) dy + \frac{1}{\sqrt{4\pi t}}x\int_{0}^{\infty}e^\frac{-(x-y)^2}{4t} dy$$
we could integrate the first term, $$\frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x-y)^2}{4t}(y-x) dy = -\frac{2t}{\sqrt{4\pi t}} e^\frac{-(x-y)^2}{4t}|_{0}^{\infty} = 0 + \frac{2t}{\sqrt{4\pi t}} e^\frac{-x^2}{4t} = \frac{\sqrt{t}}{\sqrt{\pi}} e^\frac{-x^2}{4t}$$
The second term we could substitute $y = x+z\sqrt{4t}$,
$$\frac{x}{\sqrt{\pi}}\int_{\frac{-x}{\sqrt{4t}}}^{\infty} e^{-z^2} dz\\ = \frac{x}{2}[ erf({\infty})-erf({\frac{-x}{\sqrt{4t}}})]$$
For the second part, we get  $$\frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x+y)^2}{4t}y dy = \frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x+y)^2}{4t}(y+x) dy - \frac{1}{\sqrt{4\pi t}}x\int_{0}^{\infty}e^\frac{-(x+y)^2}{4t} dy$$
Similarly, we solve first part $$\frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x+y)^2}{4t}(y+x) dy=\frac{\sqrt{t}}{\sqrt{\pi}} e^\frac{-x^2}{4t}$$
For the second part, we get $$- \frac{1}{\sqrt{4\pi t}}x\int_{0}^{\infty}e^\frac{-(x+y)^2}{4t} dy = \frac{x}{2}[ -erf({\infty})+erf({\frac{-x}{\sqrt{4t}}})]$$
Combine all the solutions, we get $$u(x,t) = 2\frac{\sqrt{t}}{\sqrt{\pi}} e^\frac{-x^2}{4t} + x(erf({\frac{-x}{\sqrt{4t}}}))$$
« Last Edit: December 18, 2015, 11:23:11 PM by Xi Yue Wang »

#### Vivian Tan

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##### Re: FE-2
« Reply #2 on: December 18, 2015, 11:07:55 PM »
Use $G_D(x-y, t) - G_D(x+y,t)$:

G_D(x-y, t) - G_D(x+y,t) = \frac{1}{\sqrt{4\pi t}}e^{\frac{-(x-y)^2}{4t}} - \frac{1}{\sqrt{4\pi t}}e^{\frac{-(x+y)^2}{4t}}

Now integrate them from 0 to infinity over y:

\frac{1}{\sqrt{4\pi t}}\int^{\infty}_0 e^{\frac{-(x-y)^2}{4t}} dy - \frac{1}{\sqrt{4\pi t}}\int^{\infty}_0 e^{\frac{-(x+y)^2}{4t}} dy

Change variables to $z_1$ and $z_2$:
$z_1 = \frac{x-y}{2\sqrt{t}}$ , $dz_1 = \frac{-1}{2\sqrt{t}}dy$,
$z_2 = \frac{x+y}{2\sqrt{t}}$ , $dz_2 = \frac{1}{2\sqrt{t}}dy$,

-\frac{1}{\sqrt{\pi}}\int^{\infty}_{\frac{x}{2\sqrt{t}}} e^{-(z_1)^2} dz_1 - \frac{1}{\sqrt{\pi}}\int^{\infty}_{\frac{x}{2\sqrt{t}}} e^{-(z_2)^2} dz_2

u(x,t) = xerf(\frac{x}{2\sqrt{t}}) + 2\sqrt{\frac{t}{\pi}}e^{-\frac{x^2}{4t}}

#### Victor Ivrii

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##### Re: FE-2
« Reply #3 on: December 20, 2015, 07:59:53 AM »
Xi Yue Wang's solution  is correct

Vivian: your solution is inconsistent with your erroneous calculations as you forgot factor $y$.

Remark. Remember $\erf$ is an odd function, $\erf(\pm \infty)=\pm 1$.

Advanced Remark. Someone tried Fourier transform and got $f=|x|\implies \hat{f}=-\frac{1}{\pi k^2}$ (but the calculations went out of hand). The real trouble with this is that F.T. of such function in the ordinary sense does not exist. It exists in the sense of distributions only and indeed $\hat{f}(k)=-\frac{1}{\pi k^2}$ outside of $0$. The r.h.e. is not a distribution in a proper sense as $\int _{-\infty}^\infty \frac{1}{k^2}\phi(k)\,dk$ does not converge in $0$ unless $\phi(0)=\phi'(0)=0$ and therefore $-\frac{1}{\pi k^2}$ should be regularized; the rule to calculate such integral should be established. The simplest way: $\ln |k|$ is a distribution and its second distributional derivative is $-k^{-2}$ regularized:
\begin{equation*}
\hat{f}(\phi)= \frac{1}{\pi}\int_{-\infty}^\infty (\ln|k|) \cdot \phi''(k)\,dk.
\end{equation*}
« Last Edit: December 20, 2015, 08:50:25 AM by Victor Ivrii »