### Author Topic: Q3-T0101  (Read 1211 times)

#### Victor Ivrii ##### Q3-T0101
« on: February 10, 2018, 06:57:13 PM »
• Find solution
\begin{equation*}
\left\{ \begin{aligned}
& u_{tt}-c^2u_{xx}=0, &&&t > 0, x > 0,  \\
&u|_{t=0}= \phi (x),   &&u_t|_{t=0}= c\phi'(x) &x > 0, \\
&(u_x+\alpha u)|_{x=0}=0,  &&&t > 0
\end{aligned}
\right.
\end{equation*}
(separately in $x>ct$, and $0<x<ct$).
• In particular, consider $\phi(x)=e^{ikx}$.

#### Jingxuan Zhang

• Elder Member
•     • Posts: 106
• Karma: 20 ##### Re: Q3-T0101
« Reply #1 on: February 11, 2018, 02:56:12 PM »
Write $u=\varphi(x+ct)+\psi(x-ct)$ then Instantly by D'Alembert' s on $x>ct$
$$\varphi=\phi,\psi=0$$
\begin{equation}u=\phi(x+ct),x>ct\end{equation}
On $0<x<ct$ solution $\varphi=\phi$ still works, whereas from boundary condition
$$\phi(-s)'+\psi'(s)+\alpha\phi(-s)+\alpha\psi(s)=0 \implies (e^{\alpha s}\psi(s))'=(e^{\alpha s}\phi(-s))'-2\alpha\phi(-s) \implies \psi(s)=\phi(-s)+2\alpha e^{-\alpha s}\int^s e^{-\alpha s'}\phi(s)\, ds'.$$
So that
\begin{equation}u=\phi(ct+x)+\phi(ct-x)+2\alpha e^{-\alpha (ct-x)}\int^{ct-x} e^{-\alpha s}\phi(s)\, ds,0<x<ct.\end{equation}
In particular, if $\phi(x)=e^{ikx}$, then
\begin{equation}u=\left\{\begin{aligned}
&e^{ik(x+ct)}&x>ct\\
&e^{ik(ct+x)}+e^{ik(ct-x)}-2\frac{ik\alpha+\alpha^2}{k^2+\alpha^2}e^{(ik-2\alpha) (ct-x)}&0<x<ct
\end{aligned}\right.
\end{equation}
Provided, indeed, $k^2+\alpha^2\neq 0$.

#### George Lu

• Jr. Member
•  • Posts: 8
• Karma: 5 ##### Re: Q3-T0101
« Reply #2 on: February 11, 2018, 06:19:59 PM »
I took a crack at the first part, but I'm not entirely confident in my steps and I was in the other section.

As always in the wave eqn, we have $u=\varphi(x+ct)+\psi$

I first solved in the region $x>ct$, where we simply have $\varphi(x)+\psi(x)=\phi(x)$ and $\varphi'(x)-\psi'(x)=\phi'(x)$, giving us $\varphi(x)=\phi(x)$ and $\psi(x)=0$.

For the region $0<x<ct$:

$\varphi'(ct)+\psi'(-ct)+\alpha\varphi(ct)+\alpha\psi(-ct)=0$

Let $t=-\frac{x}{c}$, so $\varphi'(-x)+\psi'(x)+\alpha\varphi(-x)+\alpha\psi(x)=0$

Then $\psi'(x)+ \alpha\psi(x) =-(\varphi'(-x)+\alpha\varphi(-x))$

We can see then, that $(e^{\alpha x}\psi(x))'=(e^{\alpha x}\varphi(x))'$

Integrating tells us that $\psi(x)=e^{-\alpha x}\int e^{\alpha x}\varphi(x)dx$

So $u(x,t)=\varphi(x+ct)$ for $x>ct$

And $u(x,t)=\varphi(x+ct)+e^{-\alpha x}\int_0^{ct-x} e^{\alpha x'}\varphi(x')dx'$ for $0<x<ct$

In particular, I'm not sure where the $-2\alpha\phi(-s)$ outside of the e^ derivatives in Jingxuan's solutions come from.

#### Tristan Fraser

• Full Member
•   • Posts: 30
• Karma: 11 ##### Re: Q3-T0101
« Reply #3 on: February 11, 2018, 06:34:50 PM »
I got something similar, but quite the same as Jingxan:

$$u = \phi (x+ct) , x> ct$$

On the region $0<x<ct$ , we get from the boundary condition:

$$0 = \Theta '(ct) + \Psi' (-ct) + \alpha (\Theta(ct) + \Psi(-ct))$$

Apply that $x = -ct$, and this prompts for me to rearrange and treat each side as an ODE:

$$\Psi'(x) + \alpha \Psi(x) = -(\Theta'(-x) + \alpha \Theta(-x))$$
$$(e^{\alpha x} \Psi(x) ) ' = -(\Theta'(-x) + \alpha \Theta(-x))$$
And, critically:
$$e^{\alpha x} \Psi(x)) = \Theta(-x) + \alpha \int_{0}^{-x} \Theta(x')dx' + c$$

$$\Psi(x) = e^{-\alpha x}(\Theta(-x) + \alpha \int_{0}^{-x} \Theta(x')dx' + c$$

Leading to: (note $\Theta = \phi$ in this problem)

$$u (x,t) = \phi(x+ct) + e^{\alpha (ct-x)}(\phi(ct-x) + \alpha \int_{0}^{ct-x} \Theta(x')dx' + k )$$

Examining part b: $\phi(x) = e^{ikx}$ :

$$u(x,t) = e^{ik(x+ct)} + e^{\alpha( ct - x)}( e^{ik(ct-x)} + \alpha \int_{0}^{ct-x} e^{ikx'}dx' + c)$$

$$u(x,t) = e^{ik(x+ct)} + e^{\alpha( ct - x)}( e^{ik(ct-x)} + \alpha (\frac{e^{ik(ct-x)} - 1}{ik} + c)$$

Which can be expanded into sines and cosines:

$$u(x,t) = cosk(x+ct) + i sink(x+ct) + e^{\alpha} cos(k(ct-x)) + \alpha e^{\alpha}(\frac{cosk(ct-x) + isink(ct-x) - c}{ik})$$

A few notes:

I share the same concerns as George about the $-2\alpha\phi(-s)$, and I'm not quite sure if my above solution is correct. What I do know about part b is that we are probably expected to take the mess of sines and cosines we get, and rearrange them into some form of $sin( a \pm b)$ or $cos(a \pm b)$ or we could maybe break it into $Re$ and $Im$ components?

#### Victor Ivrii ##### Re: Q3-T0101
« Reply #4 on: February 12, 2018, 06:23:22 AM »
You do not need to go to $\sin$ and $\cos$

All "mathoperators" (LaTeX terminology) like $\sin$, $\tan$, $\exp$, $\max$, $\sup$,... should be upright, so they carry escape character \: \sin

#### Victor Ivrii ##### Re: Q3-T0101
« Reply #5 on: February 12, 2018, 06:41:52 AM »
It looks like all of you got wrong (but in different ways) from ($s<0$)
\begin{align*}
&\psi'(s)+\alpha \psi(s)+ \phi'(-s)+ \alpha \phi(-s)=0\implies\\
&\psi'(s)+\alpha \psi(s)= (\phi (-s))'- \alpha \phi(-s)\implies\\
&(\psi (s)e^{\alpha s})'=(\phi (-s)e^{\alpha s})-2\alpha \phi(-s)e^{\alpha s}\implies
&\psi (s)= \phi (-s) -2\alpha e^{\alpha s}\int_0 ^s e^{\alpha s'}\phi(-s')\,ds' +C =
\phi (-s) + 2\alpha e^{\alpha s}\int_0 ^{-s} e^{-\alpha s'}\phi(s')\,ds'+C
\end{align*}
where we do change of variable $s'\mapsto -s'$. We also need a constant here so that $\psi(0^+)=\psi(0^-)$, but the former is $0$, so $C=-\phi(0)$.

« Last Edit: February 12, 2018, 06:43:52 AM by Victor Ivrii »