for us.

We can solve this problem by extending the Jacobi matrix formalism into a richer structure. This requires us to consider the general solution of Eq. , for arbitrary initial conditions. Thus, we have $$\label{jacobimat2}
\boldsymbol{\mathcal \xi}(v) = \boldsymbol{\mathcal C}(v;v{_{\text{init}}})\cdot \boldsymbol{\mathcal \xi}_{v=v{_{\text{init}}}} +
\boldsymbol{\mathcal D}(v;v{_{\text{init}}})\cdot \left.\frac{{{\rm d}}\boldsymbol{\mathcal \xi}}{{{\rm d}}v}\right\vert_{v=v{_{\text{init}}}} ,$$ as for any linear second order differential equation, solved from $v{_{\text{init}}}$ to $v$. In the following, $\boldsymbol{\mathcal C}(v;v{_{\text{init}}})$ is referred to as the [*scale matrix*]{}. It is easy to check that both the scale and Jacobi matrices satisfy the Jacobi matrix equation (\[gde2\]) but with different initial conditions: $$\label{gde2c}
\boldsymbol{\mathcal D}(v{_{\text{init}}};v{_{\text{init}}}) = \boldsymbol{0}, 
\qquad
\frac{{{\rm d}}\boldsymbol{\mathcal D}}{{{\rm d}}v}(v{_{\text{init}}};v{_{\text{init}}}) = \boldsymbol{1},$$ whereas $$\label{gde2d}
\boldsymbol{\mathcal C}(v{_{\text{init}}};v{_{\text{init}}}) = \boldsymbol{1}, 
\qquad
\frac{{{\rm d}}\boldsymbol{\mathcal C}}{{{\rm d}}v}(v{_{\text{init}}};v{_{\text{init}}}) = \boldsymbol{0}.$$

The most useful object for our problem turns out to be the $ 4\times 4$ *Wronski matrix* constructed from $\boldsymbol{\mathcal C}$ and $\boldsymbol{\mathcal D}$, $$\boldsymbol{\mathcal W}(v;v{_{\text{init}}})\equiv\left(
 \begin{array}{cc}
  \boldsymbol{\mathcal C}(v;v{_{\text{init}}}) & \boldsymbol{\mathcal D}(v;v{_{\text{init}}})\\
  \frac{{{\rm d}}\boldsymbol{\mathcal C}}{{{\rm d}}v}(v;v{_{\text{init}}}) & \frac{{{\rm d}}\boldsymbol{\mathcal D}}{{{\rm d}}v}(v;v{_{\text{init}}})
 \end{array}
 \right),$$ in terms of which the general solution (\[jacobimat2\]) reads $$\label{eq:use_wronski}
\left(
 \begin{array}{c}
  \boldsymbol{\xi}\\
  \frac{{{\rm d}}\boldsymbol{\xi}}{{{\rm d}}v}
 \end{array}
 \right)(v)=  \boldsymbol{\mathcal W}(v;v{_{\text{init}}})\cdot
 \left(
 \begin{array}{c}
  \boldsymbol{\xi}\\
  \frac{{{\rm d}}\boldsymbol{\xi}}{{{\rm d}}v}
 \end{array}
 \right)(v{_{\text{init}}}).$$ It is clear, from Eq. 