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\(r = r(x, t)\), \(q = q(x, t)\), \(\sigma_3 =

\begin{bmatrix}

1 & 0\\

0 & -1

\end{bmatrix}

\)

I have the equation

\begin{align}

\frac{\partial\mathbf{V}_{-1}^{(D)}}{\partial x} &= \frac{i}{2}\begin{bmatrix}

-(qr)_t & 0\\

0 & (qr)_t

\end{bmatrix}\\

\mathbf{V}_{-1}^{(D)} &= \alpha\sigma_3 + c\mathbb{I}\qquad (*)

\end{align}

where \(\alpha\) is a function of \(x\) and \(t\) related to \(q\) and \(r\) and

\[

\alpha_x + \frac{1}{2}i(qr)_t = 0

\]

How is \((*)\) obtained? I don't see it. I know that \(c\mathbb{I}\) is the matrix constant of integration so I am only focused on how \(\alpha\sigma_3\) comes from integrating the diagonal matrix.

I am trying to figure out the last page of

http://math.arizona.edu/~mcl/Miller/MillerLecture06.pdf

\begin{bmatrix}

1 & 0\\

0 & -1

\end{bmatrix}

\)

I have the equation

\begin{align}

\frac{\partial\mathbf{V}_{-1}^{(D)}}{\partial x} &= \frac{i}{2}\begin{bmatrix}

-(qr)_t & 0\\

0 & (qr)_t

\end{bmatrix}\\

\mathbf{V}_{-1}^{(D)} &= \alpha\sigma_3 + c\mathbb{I}\qquad (*)

\end{align}

where \(\alpha\) is a function of \(x\) and \(t\) related to \(q\) and \(r\) and

\[

\alpha_x + \frac{1}{2}i(qr)_t = 0

\]

How is \((*)\) obtained? I don't see it. I know that \(c\mathbb{I}\) is the matrix constant of integration so I am only focused on how \(\alpha\sigma_3\) comes from integrating the diagonal matrix.

I am trying to figure out the last page of

http://math.arizona.edu/~mcl/Miller/MillerLecture06.pdf

Last edited: