# Toronto Math Forum

## APM346-2018S => APM346––Home Assignments => Topic started by: Andrew Hardy on April 07, 2018, 05:38:41 PM

Title: Clever Guesses with Heat Equation
Post by: Andrew Hardy on April 07, 2018, 05:38:41 PM
Dr. Ivrii,
Jingxuan and I are studying together over the old term tests.  We're confused by your motivation for the initial guess for Fall 2016 Q2, the heat equation?

$$v(x,t)=t^{-\frac{1}{2}}e^{-\frac{x^2}{t}}$$

It ends up being quite quick, but I don't know how I could conjure up the $t^{-\frac{1}{2}}$ term on our own. Is there a method?
Title: Re: Clever Guesses with Heat Equation
Post by: Victor Ivrii on April 07, 2018, 08:16:07 PM
You mean "get the same result fast" in
http://forum.math.toronto.edu/index.php?topic=862.0 (http://forum.math.toronto.edu/index.php?topic=862.0)

We do not guess, we know that $t^{-1/2}e^{-x^2/4kt}$ satisfies  $u_t-ku_{xx}=0$. We found it in Chapter 3
Title: Re: Clever Guesses with Heat Equation
Post by: Jingxuan Zhang on April 07, 2018, 08:37:24 PM
So what if IC is not as in that question? $x^2e^{-x^3}$, say? Can we still use this or similar result?
Title: Re: Clever Guesses with Heat Equation
Post by: Victor Ivrii on April 07, 2018, 09:07:32 PM
So what if IC is not as in that question? $x^2e^{-x^3}$, say? Can we still use this or similar result?
Sometimes... but usually not. F.e. solving Cauchy problem for
\begin{align}
&u_t- ku_{xx}=0,\tag{*}\\
&u|_{t=0}=x^2e^{-ax^2}\tag{**}
\end{align}
the same way would work: again, we know that $v(x,t)=t^{-1/2}e^{-x^2/4kt}$ satisfies (*), and for  $t=t_0$ (find it) gives us $Ce^{-ax^2}$....

You need to know the regular way, cutting corners works sometimes ... but usually does not