= Solution
Freeze the coefficients at a point. Characteristic normals $(\tau,\xi)\ne(0,0)$ satisfy the homogeneous quadratic <matrix pencil> equation $\det(\tau A+\xi B)=0$. In the elementary characteristic classification, two distinct real characteristic directions give a hyperbolic system, no real directions an elliptic system, and a repeated real direction the parabolic case. An identically zero determinant polynomial is a further degeneracy, not covered by that nondegenerate classification.
When $A$ is invertible, write $C=A^{-1}B$. The finite characteristic slopes solve $\det(B-\lambda A)=0$, or
$$
\lambda^2-(\operatorname{tr}C)\lambda+\det C=0.
$$
Its discriminant $(\operatorname{tr}C)^2-4\det C$ is positive, negative, or zero in the three cases above. For the well-posed first-order evolution meaning of <hyperbolicity>, real eigenvalues must also admit a complete eigenbasis: a repeated but diagonalizable pair is non-strictly hyperbolic rather than a defective parabolic system. Distinct real roots suffice for strict hyperbolicity.
For the given <shallow water equations>, subtracting $u$ times mass conservation from momentum conservation and dividing by $h>0$ gives the primitive-variable system
$$
\begin{pmatrix}h\\u\end{pmatrix}_t+\begin{pmatrix}u&h\\1&u\end{pmatrix}\begin{pmatrix}h\\u\end{pmatrix}_x=0.
$$
The <characteristic polynomial> is $(u-\lambda)^2-h$, so
$$
\boxed{\lambda_-=u-\sqrt h,\qquad\lambda_+=u+\sqrt h}.
$$
They are real and distinct whenever $h>0$, proving strict <hyperbolicity>.
Back to article page
