Author Topic: FE Problem 7  (Read 6052 times)

Victor Ivrii

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FE Problem 7
« on: April 14, 2015, 07:49:32 PM »
Solve IVP
\begin{align*}
&u|_{t=0}=g(x)
\end{align*}
with $g(x)=x e^{-x^2/2}$.

Hint. Use Fourier transform by $x$.

Victor Ivrii

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Re: FE Problem 7
« Reply #1 on: April 17, 2015, 08:23:19 AM »
Making Fourier transform by $x$ we hade
\begin{align*}
&\hat{u}_t+\frac{k^2}{4}\hat{u}=0,\; t>0,\\
&\hat{u}|_{t=0}=\hat{g}(k).
\end{align*}
Since Fourier transform of $e^{-x^2/2}$ is $(2\pi)^{-1}e^{-k^2/2}$ we conclude that
\begin{equation*}
\hat{g}(k)=i(e^{-k^2/2})'= -(2\pi)^{-1}ike^{-k^2/2}
\end{equation*}
and
\begin{equation*}
\hat{u}(k,t)= -ike^{-k^2/2-t k^2/4}= -ik (2\pi)^{-1} e^{-k^2a^2 /2}
\end{equation*}
with $a=\sqrt{(t+2)/2)}$; then Fourier integral of $(2\pi)^{-1}e^{-k^2a^2 /2}$ is
$a^{-1}e^{-x^2 /2a^2}= \sqrt{2/(2+t)}e^{-x^2 /(2+t)}$ and finally
\begin{equation*}
u(x,t)= -\frac{\partial\ }{\partial x} \Bigl[\sqrt{2/(2+t)}e^{-x^2 /(2+t)}\Bigr]= x \Bigl(\frac{2+t}{2}\Bigr)^{-\frac{3}{2}}e^{-x^2 /(2+t)},
\end{equation*}

Victor Ivrii

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Re: FE Problem 7
« Reply #2 on: April 17, 2015, 08:23:47 AM »
Alternative solution
\begin{align*}
u(x,y)=&\frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty y\exp \bigl(-\frac{1}{t} (x-y)^2 -\frac{1}{2}y^2\bigr)\,dy=\\
&\frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty y\exp \bigl(-\frac{1}{t} x^2 +\frac{2}{t}xy  -(\frac{1}{2}+\frac{1}{t})y^2\bigr)\,dy=\\
&\frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty y\exp \bigl(-\frac{1}{t} x^2 +\frac{2}{t}xy  -\frac{t+2}{2t} y^2\bigr)\,dy=\\
&\frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty y\exp \bigl(-\frac{1}{t+2} x^2  -\frac{t+2}{2t} \bigl[y- \frac{2}{(t+2)}\bigr]^2\bigr)\,dy;
\end{align*}
plugging $y= z+\frac{2}{(t+2)}$ we get
\begin{align*}
u(x,y)=&\frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty \bigr[z+ \frac{2}{(t+2)}\bigr]\exp \bigl(-\frac{1}{t+2} x^2  -\frac{t+2}{2t} z^2\bigr)\,dz;
\end{align*}
the second factor is even by term $z$ and we can skip $z$ in the first factor (integral of odd function would be $0$);
\begin{align*}
u(x,y)=&\frac{2}{\sqrt{\pi t}(t+2)}e^{-x^2/(t+2)}  \times \int_{-\infty}^\infty  \exp \bigl(-\frac{t+2}{2t} z^2\bigr)\,dz;
\end{align*}
now integral is $\sqrt{2\pi t/(t+2)}$ (plug $z= y\sqrt{t/(t+2)}$) and
\begin{equation*}
u(x,t)= x \Bigl(\frac{2+t}{2}\Bigr)^{-\frac{3}{2}}e^{-x^2 /(2+t)}.
\end{equation*}