Past exam of the applied partial differential equations course of the University of Oxford / 2023 / 3

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 $\det (B-\lambda A)=0$, orIts discriminant $(\mathrm{tr}C{)}^2-4\det C$ 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 systemThe characteristic polynomial is $(u-\lambda {)}^2-h$, soThey are real and distinct whenever $h>0$, proving strict hyperbolicity.

Apply the Rankine-Hugoniot conditions to the conserved variables $(h,hu)$ and flux $(hu,h{u}^2+h^2/2)$. They giveFor a genuine shock, $h_1\ne 1$. Eliminating $u_1$ yields $s^2=h_1(h_1+1)/2$. Taking the positive branch requested,When $h_1=1$, the jump conditions instead force $u_1=0$ and describe a constant state; the labelled interface can then move at any speed and is not a shock. The nontrivial formulas describe a shallow-water shock into stationary unit-depth water.

For a strictly hyperbolic two-component system of conservation laws with ordered speeds ${\lambda }_{-}<{\lambda }_{+}$, a compressive family-$j$ shock has its family characteristics entering from both sides: ${\lambda }_j(U_R)<s<{\lambda }_j(U_L)$. The other family crosses the interface with the same sign of relative speed on both sides. These are the Lax shock admissibility conditions and give three incoming characteristic directions for two conservation laws.

In the present problem,Since $u_1=s-s/h_1$, we always have $u_1-\sqrt{h_1}<s$. Hence a positive-speed shock of this form cannot be a family-minus shock. For a family-plus shock the remaining strict inequalities areThe first is equivalent to $h_1(h_1+1)>2$, or $h_1>1$. The second is equivalent to $s<h_1^{3/2}$, whose square reduces to $h_1+1<2h_1^2$, again equivalent to $h_1>1$ for positive $h_1$. The minus speed on the right is $-1<s$ automatically. Thus all causal inequalities hold exactly whenDepth increases behind the right-moving shock. For $0<h_1<1$, the algebraic jump solution has the wrong characteristic orientation and is an inadmissible expansion shock; equality is the no-jump limit.