### Author Topic: TT1 = Problem 2  (Read 5241 times)

#### Victor Ivrii

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##### TT1 = Problem 2
« on: October 16, 2012, 06:27:29 PM »
Consider the initial value problem for the wave equation posed on the left half-line:
\begin{equation*}
\left\{\begin{aligned}
&u_{tt}-  u_{xx}= 0 ,\qquad&&-\infty <x< 0\\
&u (x,0) = f(x), \qquad&&-\infty < x < 0 ,\\
&u_t(x,0)= g(x), \qquad&&-\infty < x < 0.
\end{aligned}\right.
\end{equation*}
Do the initial conditions uniquely determine the solution in the region $\{ (t,x): t \in \mathbb{R}, -\infty < x < 0 \}$? Explain your answer with convincing arguments.
« Last Edit: October 16, 2012, 06:53:39 PM by Victor Ivrii »

#### Ian Kivlichan

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##### Re: TT1 = Problem 2
« Reply #1 on: October 16, 2012, 06:56:06 PM »
The initial conditions are not sufficient to uniquely define the solution - we need boundary conditions at x = 0.

Consider a counter-example to uniqueness:
1. u = 0 over x > 0 at t = 0
2. a large wave, moving in the -x direction, at t=0 (but still in the region x > 0).

Both 1. and 2. satisfy the given initial conditions, so the solution is not unique. Had we specified Dirichlet, Neumann, etc. conditions, however, one of the two cases would be impossible.
« Last Edit: October 16, 2012, 07:11:18 PM by Ian Kivlichan »

#### Victor Ivrii

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##### Re: TT1 = Problem 2
« Reply #2 on: October 16, 2012, 09:47:04 PM »
Ian, while explanation is basically correct I would like to see more convincing arguments. In particular: where solution will be defined uniquely?

#### Ian Kivlichan

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##### Re: TT1 = Problem 2
« Reply #3 on: October 16, 2012, 09:53:02 PM »
Ian, while explanation is basically correct I would like to see more convincing arguments. In particular: where solution will be defined uniquely?
With the given conditions, the I think solution is defined for -inf < x < -t, . The given conditions on u and u_t restrict it there, as any wave starting early in time would have to pass through (x,t)=(x,0). 0 < x < -t, however, does not have the required conditions for uniqueness.

#### Victor Ivrii

Also, here $t\in \mathbb{R}$. Ideal solution:
$u(x,t)= \phi(x+t)+\psi(x-t)$. Initial conditions $\phi(x)+\psi(x)=f(x)$, $\phi'(x)-\psi'(x)=g(x)$ as $x<0$ define $\phi$ and $\psi$ as $x<0$ but not as $x>0$ (we can assume with no effect to $u$ that $\phi(0)-\psi(0)=0$).
Therefore $u$ is uniquely defined as $x<-|t|$ but not as $|x|>-|t|$. Sketch would be an asset: