### Author Topic: Q3-T5102  (Read 854 times)

#### Victor Ivrii ##### Q3-T5102
« on: February 10, 2018, 06:58:27 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_{t})|_{x=0}=0,  &&&t > 0
\end{aligned}
\right.
\end{equation*}
(separately in $x>ct$, and $0<x<ct$).
• In particular, consider $\phi(x)=e^{ikx}$.

#### Tristan Fraser

• Full Member
•   • Posts: 30
• Karma: 11 ##### Re: Q3-T5102
« Reply #1 on: February 11, 2018, 01:35:32 AM »
Given:

$$u_{tt} - c^2u_{xx} = 0$$ with $t > 0, x> 0$
$$u|_{t=0} = g(x) , u_t|_{t=0} = cg'(x)$$ with $x > 0$
$$(u_x + \alpha u_t) |_{x=0} = 0$$ for $t> 0$

Generally:
$$u(x,t) = \phi(x+ct) + \psi(x-ct)$$
$$\phi(x) = g(x) - \frac{g(0)}{2}$$
$$\psi(x) = \frac{g(0)}{2}$$

It follows for the nicer region ($x > ct$) we have:

$$u(x,t) = \phi(x+ct) + \psi(x-ct) = g(x+ct) -\frac{g(0)}{2} + \frac{g(0)}{2} = g(x+ct)$$

While for the case of ($0<x<ct$) means that we have to investigate the behaviour of \psi($\gamma$) with $\gamma = x - ct < 0$

Applying the boundary condition of  $(u_x + \alpha u_t) |_{x=0} = 0$

Resulting in :

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

Rearranging, and applying the substitution of $x = -ct$ this gives us:

$$\psi ' (x) = \phi ' (-x) \frac{ (1 + \alpha)}{\alpha - 1 }$$

This is where I usually have trouble with these sorts of problems, where I don't know where/if I am allowed to integrate both sides/ use the original expressions for $\phi$ and $\psi (x)$ above. Although I do know that I should try to manipulate it into the form of an ODE.

Tentatively, I'd start by taking $\phi'(x) - \psi'(x) = g'(x)$  rearranging to:
$$\phi'(x) = g'(x) + \psi'(x)$$

the issue is that substituting this expression into the above results in results in an expression of $\psi'(x)$ and $\psi'(-x)$ , so clearly this tentative step is incorrect, but I am unsure of what to do next.

b. Will be discussed later

#### Elliot Jarmain

• Jr. Member
•  • Posts: 6
• Karma: 4 ##### Re: Q3-T5102
« Reply #2 on: February 11, 2018, 08:46:36 AM »
Based on examples from the online textbook, I am pretty sure we are allowed to integrate at that stage. Also Example 2 b from Section 2.6 seems to discard the constants of integration so I will follow that convention. Here is my solution to part 1:

We know $u(x,t)$ is of the form $u(x,t) = \varphi(x+ct)+\psi(x-ct)$.
For $x>ct$ we use the initial conditions:
\begin{align}
u|_{t=0}= \phi (x)
&\implies \varphi(x)+\psi(x) = \phi(x)  \\
u_t|_{t=0}= c\phi'(x)
&\implies \varphi'(x)-\psi'(x)=\phi'(x)
\end{align}
Therefore $\varphi(x)=\phi(x)$ and $\psi(x)=0$ when $x>0$. So
\begin{equation}
\text{when } x>ct
\end{equation}
For $0<x<ct$ we use the boundary condition:
\begin{align}
(u_x+\alpha u_t)|_{x=0}=0
&\implies \varphi'(ct) + \psi'(-ct)
+ \alpha c\varphi'(ct) -\alpha c\psi'(-ct) = 0 \\
&\implies \frac{1}{c}\varphi(ct) -                              \frac{1}{c}\psi(-ct)
+ \alpha \varphi(ct) +\alpha\psi(-ct) = k
\end{align}
Letting $t=-\frac{x}{c}$ and rearranging yields
\begin{align}
\psi(x) &= - \left(
\frac{\alpha c + 1}{\alpha c -1} \right)
\text{for } x<0 \\
&= - \left(
\frac{\alpha c + 1}{\alpha c -1} \right)
\phi(-x) + k
\end{align}
So
\begin{equation}
u(x,t) = \phi(x+ct)- \left(
\frac{\alpha c + 1}{\alpha c -1} \right)
\text{when } 0<x<ct
\end{equation}
(we set $k=0$ so that $u$ is continuous at $x=ct$)

Therefore,
\begin{equation}
u(x,t) =
\left\{ \begin{aligned}
&\phi(x+ct), &&x>ct \\
&\phi(x+ct) - \left(
\frac{\alpha c + 1}{\alpha c -1} \right)
\phi(ct-x), &&0<x<ct
\end{aligned}
\right.
\end{equation}
« Last Edit: February 11, 2018, 08:59:14 AM by Elliot Jarmain »

#### Victor Ivrii ##### Re: Q3-T5102
« Reply #3 on: February 11, 2018, 09:07:01 AM »
Indeed, in this example we can even get rid of the integral. Please tell me, what is a condition to $\alpha$ for the problem to be well-posed

#### Elliot Jarmain

• Jr. Member
•  • Posts: 6
• Karma: 4 ##### Re: Q3-T5102
« Reply #4 on: February 11, 2018, 09:13:10 AM »
For the problem to be well-posed, $\alpha \neq \frac{1}{c}$.