### Author Topic: P2 Night  (Read 1358 times)

#### Victor Ivrii

• Elder Member
• Posts: 2555
• Karma: 0
##### 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}\\
\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

• Full Member
• Posts: 30
• Karma: 11
##### 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

• Elder Member
• Posts: 2555
• Karma: 0
##### 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

u(x,t)=
3\iint_{\Delta (x,t)}\, e^{-\xi^2}\, d\xi d\tau,
\label{2-3}

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]