 Solution

ID: past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2010/b/5/a/solution

Write the system as

U_{x} + A (U) U_{y} = 0

with

U = (u, v)^{T}

and

A = (v 1 u v), det (A - λ I) = (v - λ)^{2} - u .

λ_{\pm} = v \pm u

, so it is strictly hyperbolic precisely when $u > 0$ . At

u = 0

they coincide and the matrix is not diagonalizable; for

u < 0

the speeds are nonreal.

For

u > 0

R_{\pm} = v \pm 2 u

and

D_{\pm} = \partial_{x} + (v \pm u) \partial_{y}

. Substitution from the two differential equations gives

D_{\pm} R_{\pm} = v_{x} + (v \pm u) v_{y} \pm u^{- 1/2} [u_{x} + (v \pm u) u_{y}] = (- u_{y} \pm u v_{y}) \pm u^{- 1/2} (\pm u u_{y} - u v_{y}) = 0.

Therefore

v \pm 2 u is constant on curves \frac{d y}{d x} = v \pm u .

These are the two Riemann invariants; their constancy applies within smooth regions, not across a shock.

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