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I've been asked to find the Riemann Invariants for the system: $$ \begin{pmatrix} \cos(v) & 0 \\ 0 & \cos(v) \end{pmatrix} \begin{pmatrix} u_x \\ v_y \end{pmatrix} + \begin{pmatrix} \sin(v) & -1 \\ -1 & \sin(v) \end{pmatrix} \begin{pmatrix} u_y \\ v_y \end{pmatrix} = 0$$

It's easy to show that the two families of characteristic projections satisfy:

$$\frac{dy}{dx} = \tan(v) \pm \sec(v)$$ and that along these curves we have, for $\underline{u} = (u,v)$:

$$(1,\mp1) (\cos(v)\frac{\partial \underline{u}}{\partial x} +(\sin(v)\pm1)\frac{\partial \underline{u}}{\partial y}) = 0 $$

I'm only really having trouble with the last step, which is finding the Riemann invariants from this (i.e. the $R_\pm$ such that the above expression can be written as $\frac{d}{dx}R = 0$.)

Any help in seeing what these Riemann invariants are would be appreciated.

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  • $\begingroup$ Considering you found the eigenvalues of the matrix, did you calculate the left eigenvectors at all? $\endgroup$ – Mattos Nov 10 '17 at 15:37
  • $\begingroup$ @Mattos yes the left eigenvectors are the $(1,\pm 1)$ vectors $\endgroup$ – Fahrenheit997 Nov 10 '17 at 15:56

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