### Author Topic: Halloween Challenge 6  (Read 228 times)

#### Victor Ivrii

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##### Halloween Challenge 6
« on: October 21, 2016, 12:06:16 PM »

#### XinYu Zheng

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##### Re: Halloween Challenge 6
« Reply #1 on: November 01, 2016, 02:52:20 PM »
Suppose for a moment that $t>0$. We know that the heat equation $u_t=ku_{xx}$ has a self similar solution of $u(x,t)=(4\pi kt)^{-1/2}e^{-\frac{x^2}{4kt}}$. For this equation, we just make the change $k\to ik$ and obtain
$$u(x,t)=(4\pi i kt)^{-1/2}e^{\frac{i x^2}{4kt}}$$
Note that $i^{-1/2}=e^{-(\log i)/2}$. Taking the principle branch of the complex logarithm where the argument lies in $[-\pi, \pi)$, we have $i^{-1/2}=e^{-i \pi/4}$. Thus for $t>0$ we have the solution
$$u(x,t)=(4\pi kt)^{-1/2}e^{\frac{i x^2}{4kt}-\frac{i \pi}{4}}$$
Now, if $t<0$, note that $u(ix,-t)$ satisfies the original ODE. So we make the change $x\to ix$ and $t\to -t$ to obtain the solution for $t<0$:
$$u(x,t)=(-4\pi kt)^{-1/2}e^{\frac{i x^2}{4kt}-\frac{i \pi}{4}}$$
Putting this together we have
$$u(x,t)=(4\pi k|t|)^{-1/2}e^{\frac{i x^2}{4kt}-\frac{i \pi}{4}}$$
« Last Edit: November 01, 2016, 03:06:29 PM by XinYu Zheng »

#### Victor Ivrii

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##### Re: Halloween Challenge 6
« Reply #2 on: November 02, 2016, 06:14:25 AM »
$\renewcommand{\Re}{\operatorname{Re}}$
We assume that $k>0$. You are correct that we need to plug $k:= ik$ but this should be done carefully (this is important only for $(4\pi k t)^{-\frac{1}{2}}$).

As $t>0$ we can consider complex $k\in \mathbb{C}$, with $\Re k>0$ and therefore we need to plug $k:=ik+0$ and then $k^{-\frac{1}{2}}\mapsto k^{-\frac{1}{2}} e^{-i\pi/4}$.

As $t<0$ we can consider complex $k\in \mathbb{C}$, with $\Re k<0$ and therefore we need to plug $k:=ik-0$ and then $k^{-\frac{1}{2}}\mapsto k^{-\frac{1}{2}} e^{i\pi/4}$.

So, you got a correct answer for $t>0$, for $t<0$ the factor is a bit different:

u= (4\pi k |t|)^{-\frac{1}{2}} e^{\frac {ix^2}{4kt} \mp \frac{i\pi }{4}}\qquad \text{for}\; \pm t>0.

#### XinYu Zheng

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##### Re: Halloween Challenge 6
« Reply #3 on: November 02, 2016, 12:55:18 PM »
Ah I see. Then there is a typo in http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter5/S5.3.html#sect-5.3.2, where the sign on $ix^2/(4kt)$ in the exponent should be positive, and the sign on $i\pi/4$ should be $\mp$ not $\pm$.

Edit: Wait a second. Why does switching from $ik+0$ to $ik-0$ change the argument of the complex number? $ik+0$ and $ik-0$ have the same angle in the complex plane.
« Last Edit: November 02, 2016, 02:35:41 PM by XinYu Zheng »

#### Victor Ivrii

No: because the cut goes along $\Re k=0$: for $\Re k \gtrless 0$ problem is wellâ€“posed as $t\gtrless 0$