Author Topic: Q3-T5102  (Read 854 times)

Victor Ivrii

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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

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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:
$$ u(x,t) = \phi(x+ct) + \psi(x-ct) $$
The initial conditions lead to:
$$ \phi(x) = g(x) - \frac{g(0)}{2} $$
$$ \psi(x) = \frac{g(0)}{2} $$

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

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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}
    u(x,t) = \phi(x+ct), \quad
    \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)
    \varphi(-x) + k, \quad
    \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)
    \phi(ct-x), \quad
    \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

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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

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