Author Topic: P2 Night  (Read 1293 times)

Victor Ivrii

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P2 Night
« on: February 15, 2018, 08:19:03 PM »
$\newcommand{\erf}{\operatorname{erf}}$
Find solution $u(x,t)$ to
\begin{align}
&u_{tt}-9u_{xx}=18 e^{-x^2},\tag{1}\\
&u|_{t=0}=0, \quad u_t|_{t=0}=0.\tag{2}
\end{align}

Hint: Change order of integration over characteristic triangle.
Use $\erf(x)=\frac{2}{\sqrt{\pi}}\int _0^x e^{-z^2}\,dz$.
« Last Edit: February 15, 2018, 09:31:01 PM by Victor Ivrii »

Tristan Fraser

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Re: P2 Night
« Reply #1 on: February 19, 2018, 11:55:31 AM »
With the initial conditions implying:

$ g(x) = h(x) = 0 $, and $c = 3 $.
This means that D'Alembert's formula simplifies, from:

$$ u(x,t) = \frac{1}{2} (g(x+ct)+g(x-ct)) + \frac{1}{2c} \int_{x-ct}^{x+ct} h(y)dy + \frac{1}{2c}\int_{0}^{t} \int_{x-c(t-t')}^{x+c(t-t')} f(x',t')dx'dt' $$

Into:

$$ u(x,t) =  \frac{1}{6}\int_{0}^{t} \int_{x-3(t-t')}^{x+3(t-t')} f(x',t')dx'dt' $$

As suggested, rearranging for the order of integration would be convenient here. For simplicity, I also broke up the characteristic triangle into two smaller triangles.

For the first one:

$ 0 < t' < x \frac{x'-x+3t}{3} $ and $ x-3t < x' < x$

And the second:


$ 0 < t' < x \frac{-x'+x+3t}{3} $ and $ x < x' < x+3t$

This breaks our integral up into the following:

$$ (1) = \frac{1}{6} \int_{x-3t}^{x} \int_{0}^{\frac{x' - x + 3t}{3}} 18e^{-x'^2}dt'dx' $$
$$ (2) = \frac{1}{6} \int_{x}^{x+3t} \int_{0}^{\frac{-x' + x + 3t}{3}} 18e^{-x'^2}dt'dx' $$
Where $ u(x,t) = (1)+(2) $

Starting with (1), we first integrate to get:

$$ (1) = \frac{1}{6} \int_{x-3t}^{x} \frac{x' - x +3t}{3} 18e^{-x'^2} dx' $$
$$ (1) = 3 \int_{x-3t}^{x} (3t-x)e^{-x'^2}dx'  + 3 \int_{x-3t}^{x}  x'e^{-x'^2} dx' $$

The first integral can be resolved to error functions, the second requires a simple substitution to solve:
Let

$ u = x^2 $ then $ du = 2xdx $. Thus, for second term $\int_{(x-3t)^2}^{x^2} \frac{3}{2} e^-u du  = \frac{3}{2} ( e^{-(x-3t)^2} - e^{-(x)^2}) $

Putting it together, we get that :


$$ (1) = 3\sqrt{\pi} (3t-x) (erf(x) - erf(x-3t)) + \frac{3}{2} ( e^{-(x-3t)^2} - e^{-(x)^2}) $$

Likewise, for (2) we should get:

$$ (2) = 3\sqrt{\pi} (3t+x) (erf(x+3t) - erf(x)) - \frac{3}{2} ( e^{-(x)^2} - e^{-(x+3t)^2}) $$

Then

$$ u(x,t) = 3 \sqrt{\pi} (3t(erf(x+3t) - erf(x-3t)) - 2xerf(x) + x(erf(x+3t)+erf(x-3t)) - 3e^{-x^2} + \frac{3}{2} (e^{-(x-3t)^2} + e^{-(x-3t)^2} $$



Victor Ivrii

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Re: P2 Night
« Reply #2 on: February 22, 2018, 07:32:19 AM »
A bit more polished:

$\newcommand{\erf}{\operatorname{erf}}$
By D'Alembert formula
\begin{equation}
u(x,t)=
3\iint_{\Delta (x,t)}\, e^{-\xi^2}\, d\xi d\tau,
\label{2-3}
\end{equation}
where $\Delta (x,t)$ is bounded by $\tau=0$, $x-\xi-3(t-\tau)=0$, $x-\xi +3(t-\tau)=0$.

Then the double integral becomes
\begin{align*}
&3 \int_{x-3t}^x \Bigl(\int_0^{t+\frac{1}{3}(\xi-x)}d\tau \Bigr)e^{-\xi^2}\,d\xi+
3 \int^{x+3t}_x \Bigl(\int_0^{t-\frac{1}{3}(\xi-x)}d\tau \Bigr)e^{-\xi^2}\,d\xi=\\[2pt]
&\int_{x-3t}^x (3t-x+\xi )e^{-\xi^2}\,d\xi + \quad \qquad \int^{x+3t}_x (3t+x-\xi)e^{-\xi^2}\,d\xi=\\
&(3t-x) \int_{x-3t}^x e^{-\xi^2}\,d\xi -\frac{1}{2}e^{-\xi^2}\bigr|_{\xi= x-3t}^{\xi=x} +
(3t+x) \int_x^{x+3t} e^{-\xi^2}\,d\xi +\frac{1}{2}e^{-\xi^2}\bigr|^{\xi= x-3t}_{\xi=x} =\\[4pt]
&(3t-x)\Bigl(\erf(x)-\erf(x-3t)\Bigr)+ (3t+x)\Bigl(\erf(x+3t)-\erf(x)\Bigr)+\\[4pt]
& \qquad\qquad\qquad\qquad\qquad\qquad\qquad\frac{1}{2}\Bigl(e^{-(x-3t)^2}+e^{-(x+3t)^2}-2e^{-x^2}\Bigr).
\end{align*}