# Problem related to wave equation

I am stuck with the following problem:

If $u(x,t)$ satisfies the wave equation:
$u_{tt}=c^2u_{xx}, x \in \Bbb R,t>0$ ,with initial conditions

$u(x,0)=\begin{cases}\sin\dfrac{\pi x}{c}&0\leq x\leq c\\0&\text{elsewhere}\end{cases}$

and $u_t(x,0)=0$ $\forall x$ , then for a given $t>0$ ,

1. there are values of $x$ at which $u(x,t)$ is discontinuous

2. $u(x,t)$ is continuous,but $u_x(x,t)$ is not continuous

3. $u(x,t)$ and $u_x(x,t)$ are continuous,but $u_{xx}(x,t)$ is not continuous

4. $u(x,t)$ is smooth for all $x$ .

I am studying the Cauchy problem for the wave equation $n=2$ ; $$\begin{cases}u_{tt}=\alpha^{2} u_{xx}, x \in\mathbb{R}, t>0\\[8pt] u(x,0)=f(x), x\in\mathbb{R}\\[8pt] u_{t}(x,0)=g(x), x\in\mathbb{R} \end{cases}$$

By d´Alembert's formula we know that

$u(x,t)=\dfrac{f(x+\alpha t)+f(x-\alpha t)}{2} +\dfrac{1}{2\alpha} \displaystyle\int^{x+\alpha t}_{x-\alpha t}g(s)ds=\dfrac{1}{2}[f(x+ct)+f(x-ct)]$ , since here $\alpha=c$ , $g(s)=0$ .

Now I am not sure how to progress hereon. Can someone point me in the right direction? Thanks in advance for your time.

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HINT: The d´Alembert's formula essentially says that the discontinuities of the initial condition $f(x)$ and $g(x)$ will propagate at the speed of $c$, as discontinuities of $u(x,t)$ and $u_x(x,t)$ respectively. And try to translate this physical and heuristic statement to mathematical language. –  Shuhao Cao Apr 24 '13 at 4:16

The solution u(x,t) is just two half-height replicas of the initial condition moving in opposite directions (see the sketch). From this it is obvious that u(x,t) is continuous but its derivative is not - so the right choice is 2.

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You got this far using d'Alembert's formula:

$$u(x,t)=\frac{f(x+ct)+f(x-ct)}{2} = \frac{u(x+ct,0)+u(x-ct,0)}{2}$$ This says that the total wave is the sum of two pulses propagating away from the origin with speed $c$. Each pulse has a spatial shape that is the same as the initial shape. Let's try evaluating the various statements.

there are values of $x$ at which $u(x,t)$ is discontinuous

Since $u(x,t)$ is made up of two pulses propagating in different directions and each one has the shape of the initial condition, this boils down to the continuity of the initial condition. The initial condition is a truncated half-cycle of a sinusoid. For continuity of a pulse like this, all you have to check is that the function is zero at the endpoints where it's truncated. In this case, $\sin(\pi x/c)$ goes to zero at $x=0$ and $x=c$, so $u(x,t)$ is continuous as a function of x.

$u(x,t)$ is continuous,but $u_x(x,t)$ is not continuous

Let's evaluate $u_x(x,t)$ $$u_x(x,t) = \frac{f'(x+ct) + f'(x-ct)}{2}$$ So is the initial condition's derivative continuous at the truncated endpoints? Well on the interval where it's nonzero, $f'(x) = \frac{\pi}{c}\cos(\pi x/c)$, $f'(0) = \pi/c$, $f'(c) = -\pi/c$. But outside the interval $f'(x)=0$, so clearly, there is a mismatch in the derivative at the pulse edges. This option is looking good.

$u(x,t)$ and $u_x(x,t)$ are continuous,but $u_{xx}(x,t)$ is not continuous

We already identified that $u_x(x,t)$ is discontinuous, so you can discount this one. You can calculate $u_{xx}(x,t)$ for good measure though. It boils down to deciding if the second derivative of the pulse shape is zero at the ends. In this case, $f''(x)=-\frac{\pi ^2 \sin \left(\frac{\pi x}{c}\right)}{c^2}$, $f''(0)=f''(c)=0$, so it turns out that $u_{xx}(x,t)$ is indeed continuous in $x$.

$u(x,t)$ is smooth for all $x$

We can rule out smoothness since we found a "kink", a first derivative discontinuity. Having thoroughly examined all possibilities, the second one is the one that is true.

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$\because$ according to http://mathworld.wolfram.com/BoxcarFunction.html, we can rewrite $u(x,0)=\begin{cases}\sin\dfrac{\pi x}{c}&0\leq x\leq c\\0&\text{elsewhere}\end{cases}$ as $u(x,0)=\prod_{0,c}(x)\sin\dfrac{\pi x}{c}$

$\therefore$ you can directly substitute it into the d´Alembert's formula:

$u(x,t)=\dfrac{1}{2}\biggl(\prod_{0,c}(x+ct)\sin\dfrac{\pi(x+ct)}{c}+\prod_{0,c}(x-ct)\sin\dfrac{\pi(x-ct)}{c}\biggr)$

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