Toronto Math Forum
APM3462018S => APM346––Home Assignments => Web Bonus Problems => Topic started by: Victor Ivrii on January 25, 2018, 06:47:02 PM

As we know, solution to
\begin{align}
&u_t u_{xx}=0, \label{A}\\
&u_{t=0}=\delta(x)\label{B}
\end{align}
is
\begin{equation}
\frac{1}{\sqrt{4\pi t}}e^{x^2/4t}
\label{C}
\end{equation}
where initial condition currently is understood as $u(x,t)\to 0$ as $t\to +0$, $x\ne 0$; $u(0,t)\to \infty$ as $t\to +0$ and $\int_{\infty}^\infty u(x,t)\,dx=1$.
a. Consider 1D "radioactive cloud" problem:
\begin{align}
&u_t + v u_xu_{xx}+\beta u=0, \label{D}\\
&u_{t=0}=\delta(x)\label{E},
\end{align}
where $v$ is a wind velocity, $\beta$ shows the speed of "dropping on the ground".
Hint: Reduce to (\ref{A})(\ref{B}) by $u= ve^{\beta t}$ and $x=y+vt$, use (\ref{C}) for $v$ and then write down $u(x,t)$.
b. Find "contamination level" at $x$
\begin{equation}
D(x)=\beta \int _0^\infty u(x,t)\,dt.
\label{F}
\end{equation}
Hint: by change of variables $t= y^2$ with appropriate $z$ reduce to calculation of
\begin{equation}
\int \exp(ay^2by^{2})\,dy
\label{G}
\end{equation}
and calculate it using f.e. https://www.wolframalpha.com/ (https://www.wolframalpha.com/) with input
int_0^infty exp (ay^2b/y^2)dy
(you may need to do it few times)
c. Later

a. Following the hint let $\tilde{u}(y,t):=u(y+vt,t)= v(y+vt,t)e^{\beta t}=\tilde{v}(y,t)e^{\beta t}$ then
$$(4) \implies \tilde{u}_t \tilde{u}_{yy}+\beta \tilde{u} = 0 \implies \tilde{v}_t \tilde{v}_{yy} = 0 \implies \tilde{v} = \frac{1}{\sqrt{4\pi t}}e^{y^2/4t} \implies u = \frac{1}{\sqrt{4\pi t}}e^{((xvt)^2+4\beta t^2)/4t}$$
b. But I really don't know how to use that program!

Rewrite $[(xv t)^2 +4\beta t^2]/4t$ as $at +bt^{1}$ and plug $t=y^2$ in the integral.
In Wolfram Alpha there is a single window to enter the code, which is the similar to LaTeX ... and I provided you with the code...
You don't know "copypaste"? :D

Okay let me confess: it was really due to shame of not being able to compute that integral by hand that I did not attempt to Wolfram it. From there I have:
$$D(x)=\frac{\beta}{\sqrt{v^2+4\beta}}\exp(x(v/2\sqrt{v^2/2+\beta}))$$