Why do the pulses move like that?

[Goklayeh]

Hello everybody! I have a really silly question concerning wave equation: consider the problem
$$\left\{ \begin{matrix} u_{tt} &=& u_{xx} & x \in \mathbb{R}\\ u(x,0) &=& 0 & \\ u_t(x,0) &=& x(1-x)\chi_{\left[0,1\right]}(x)& \end{matrix} \right.$$
the solution is given by d'Alembert's formula
$$u(x,t) = \int_{x-t}^{x+t}{y(1-y)\chi_{\left[0,1\right]}(y) \mathrm{d}y} = \int_{\mathbb{R}}{y(1-y)\chi_{\left[0,1\right]\cap \left[x-t,x+t\right]}(y) \mathrm{d}y}$$
Now, it's clear that $$\forall x_0 \in \mathbb{R}\:\: \exists t_0$$ s.t. $$u(x_0,t) \ne 0\:\: \forall t > t_0$$. For instance, fix $$x_0 = -1$$. Then, for all $$t \ge 2$$, we have $$u(-1,t) = \int_0^1{(y - y^2)\mathrm{d}y} = \frac{1}{6}\ne 0$$. But, physically, this means that the wave passes through $$x_0$$ definitively, i.e. for all enough big times. How is this possible? Or, where I'm wrong?
Thank you for your attention

 

[LCKurtz]

I think you are missing a factor of 1/2 in your solution, but never mind that. Your solution has the form

u(x,t) = f(x+t)χ(x+t) - f(x-t)χ(x-t)

This describes one pulse moving to the left and the other moving to the right as t increases. If for example you look at x = 10 along the string and t > 0, χ(10+t) will always be 0 and χ(10-t) will only be 1 when t is between 9 and 10. During that 1 second time you will have the string displaced by f(10-t)*1 as the pulse moves by to the right.

Similarly at x = -9, χ(-9+t) will be 1 when t is between 9 and 10 and you have the pulse going the other direction.