Wave equation for lightcone coordinate $X^-$ A quick question from Polchinski volume.1 : He claims in p.20 that the worldsheet lightcone coordinates $X^\pm$ also (i.e. in addition to the transverse coordinates $X^i$) satisfy the wave-equation. Of course, for $X^+$, it is trivial since we fix  the lightcone gauge :
$$
\tau=X^+
$$ implying $\partial_\tau^2 X^+ = c^2 \partial_\sigma^2 X^+ = 0$.
How can we show that the wave equation for the coordinate $X^-$ :
$$
\partial_\tau^2 X^- = c^2 \partial_\sigma^2 X^- \quad,\quad c \equiv \frac{l}{(2\pi\alpha')p^+} 
$$ is also implied by the Hamiltonian EOMs which are :
\begin{eqnarray}
\frac{H}{p^+} &=& \partial_\tau x^- \\ 0 &=& \partial_\tau p^+ \\ c(2\pi \alpha') \Pi^i &=& \partial_\tau X^i \\ \partial_\tau \Pi^i &=& \frac{c}{(2\pi\alpha')} \partial_\sigma^2 X^i
\end{eqnarray}
and here $x^-(\tau)$ is the `mean-value' of $X^-$ :
$$ x^-(\tau) = \frac{1}{l} \int_0^l d\sigma X^-(\tau,\sigma) 
$$
I think one needs to compute $\partial_\tau H$ to obtain the expression for $\partial_\tau X^-$, but it doesn't seem to give the desired result. Perhaps, I am missing some obvious point.
 A: \begin{equation}
\partial_{\tau}x^{-}(\tau)=\frac{H}{p^{+}};\hspace{2mm}H=\frac{l}{4\pi\alpha^{\prime}p^{+}}\displaystyle\int_{0}^{l}d\sigma\big(2\pi\alpha^{\prime}\Pi^{i}\Pi^{i}+\frac{1}{2\pi\alpha^{\prime}}\partial_{\sigma}X^{i} \partial_{\sigma} X^{i} \big)
\end{equation}
\begin{equation}
\begin{split}
\therefore \hspace{3mm}\partial^{2}_{\tau}x^{-}(\tau)&=\frac{1}{p^{+}}\partial_{\tau}H\\&=\frac{l}{4\pi\alpha^{\prime}(p^{+})^{2}}\displaystyle\int_{0}^{l}d\sigma\hspace{1mm}\big[4\pi\alpha^{\prime}\Pi^{i}\partial_{\tau}\Pi^{i}+\frac{1}{\pi\alpha^{\prime}}(\partial_{\sigma}\partial_{\tau}X^{i})\partial_{\sigma}X^{i} \big]\\&=\frac{l}{(2\pi\alpha^{\prime}p^{+})^{2}}\displaystyle\int_{0}^{l}d\sigma\big[\partial_{\tau}X^{i}\partial^{2}_{\sigma}X^{i}+(\partial_{\sigma}\partial_{\tau}X^{i})\partial_{\sigma}X^{i}  \big]\\&=\frac{l}{(2\pi\alpha^{\prime}p^{+})^{2}}\big[\displaystyle\int_{0}^{l}d\sigma\hspace{2mm}\partial_{\tau}X^{i}\partial^{2}_{\sigma}X^{i} +\partial_{\sigma}X^{i}\partial_{\tau}X^{i}\Bigr|_{0}^{l}-\displaystyle\int_{0}^{l}d\sigma\hspace{2mm}\partial_{\tau}X^{i}\partial^{2}_{\sigma}X^{i} \big]\\&=0
\end{split}
\end{equation}
using the boundary condition given in eqn. (1.3.15).
And, as $ x^{-} $ is a function of $\tau$ only,we have $$ \partial^{2}_{\sigma}x^{-}=0 $$
\begin{equation}
\therefore\hspace{13mm} \partial^{2}_{\tau}x^{-}=c^{2}\partial^{2}_{\sigma}x^{-}=0
\end{equation}
Now we can use the expression of $x^{-}(\tau)$ given in terms of $X^{-}(\tau,\sigma)$ by eqn. (1.3.12a)to get,
\begin{equation}
\begin{split}
\frac{1}{l}\partial^{2}_{\tau}\displaystyle\int_{0}^{l}d\sigma X^{-}(\tau,\sigma)=\frac{c^{2}}{l}\partial^{2}_{\sigma}\displaystyle\int_{0}^{l}d\sigma X^{-}(\tau,\sigma)=0
\end{split}
\end{equation}
or,
\begin{equation}
\partial^{2}_{\tau}X^{-}=c^{2}\partial^{2}_{\sigma}X^{-}
\end{equation}
