 Solution

ID: past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2017/6/b/solution

\dot{δ x} = f (x + δ x) - f (x) = J (t) δ x + O (∥ δ x ∥^{2}), J_{ij} = \frac{\partial f _{i}}{\partial x _{j}} .

Over a sufficiently short interval, the tangent-linear propagator is

δ x (t + δ t) = A δ x (t), A = I + J δ t + O (δ t^{2}) .

Consequently

A^{T} A = I + (J + J^{T}) δ t + O (δ t^{2}) .

S = J + J^{T}

has

S e = λ e

and the perturbation is initially parallel to

e

, its squared Euclidean norm obeys

∥ δ x (t + δ t) ∥^{2} = (1 + λ δ t) ∥ δ x (t) ∥^{2} + O (δ t^{2} ∥ δ x (t) ∥^{2}) .

Positive

λ

means instantaneous norm-squared growth, negative

λ

decay, and zero

λ

no first-order change. The corresponding norm growth rate is

λ /2

, not

λ

. These are eigenvalues of the symmetric part governing the chosen error metric, not generally the eigenvalues of the nonsymmetric Jacobian matrix or the long-time Lyapunov exponents. The direction and the reference Jacobian matrix evolve during a finite forecast.

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