Toronto Math Forum
APM346-2018S => APM346--Tests => Quiz-3 => Topic started by: Victor Ivrii on February 10, 2018, 06:58:27 PM
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- 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 $\phi(x)=e^{ikx}$.
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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
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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}
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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
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For the problem to be well-posed, $\alpha \neq \frac{1}{c}$.