$\H$, $\E_1$, $\E_2$ constitute a compatible Hamiltonian triple. These operators can be combined to obtain a new pair of compatible Hamiltonian operators: $$\label{mkdv-peakon-ops}
\wtil\H = \H - \E_1=-D_x + D_x^3, 
\quad
\wtil\E= \E_2=-D_xvD\inv_x vD_x .$$

The following steps are used to derive a peakon equation from this Hamiltonian pair . First, observe that the operator $\wtil\H$ has the factorization $$\label{mkdv-Delaops}
\wtil\H = -D_x\Delta = -\Delta D_x,
\quad
\Delta = 1- D_x^2$$ where $\Delta=\Delta^*$ is a symmetric operator. Second, introduce a potential variable $u$ in terms of $v$, $$\label{mkdv-potential}
v = \Delta u$$ where these variables satisfy the variational relation $$\label{mkdv-varders}
\Delta\frac{\delta}{\delta v} =\frac{\delta}{\delta u} .$$ Next, consider the flow defined by $x$-translations on $v$, $P^{(0)} = -v_x$. This root flow inherits a natural bi-Hamiltonian structure with respect to the two Hamiltonian operators . The first Hamiltonian structure arises from expressing $-v_x = \wtil\H(\Delta\inv v)$ with $\Delta\inv v= \delta \wtil H/\delta v$, so then $$-v_x = \wtil\H(\delta \wtil H/\delta v)$$ with $v= \Delta(\delta \wtil H/\delta v) = \delta \wtil H/\delta u$ through the variational relation , where the Hamiltonian is given by $$\label{mkdv-peakon-H}
\wtil H = \int_{-\infty}^{\infty} \tfrac{1}{2} uv\,dx
= \int_{-\infty}^{\infty} \tfrac{1}{2}(u^2 +u_x^2)\,dx$$ using expression and integrating by parts. The second 