Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2017 / 6 / c / Solution

For the instantaneous Lorenz error growth at a zero-y crossing, direct differentiation of the Lorenz system givesAt a crossing of $y=0$ its symmetric part is particularly simple:Thus a perturbation initially in the $z$ direction has norm-squared rate $-2b$ and initially decays in norm at rate $b$. The $x$-dependent couplings between $y$ and $z$ are skew-symmetric at the crossing and do not contribute to the instantaneous total norm derivative. They can rotate the perturbation later: a pure $\delta z$ immediately induces $\dot{\delta y}=-x\delta z$. It is therefore incorrect to claim that such an error remains an independently decaying $z$ mode along the complete trajectory.

For the $(x,y)$ block, the trace is $-2(\sigma +1)<0$ and the determinant is $4\sigma -(\sigma +r-z{)}^2$. A negative determinant gives one positive and one negative eigenvalue. Equivalently,In particular the requested sufficient lower-height condition isThere is also a second growth range $z>\sigma +r+2\sqrt{\sigma }$; it must not be lost by removing the absolute value. At either endpoint the largest rate is zero. Between them both block eigenvalues are negative. At a crossing in either growing range, some planar error directions have positive instantaneous exponential rate. This statement is local in time: a single instantaneous positive eigenvalue of $S$ does not prove sustained exponential error growth over an evolving Lorenz trajectory.