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In vector form the system is $U_x+A(U)U_y=C(U)$, withFor $v>0$, its two real distinct eigenvalues are $\boxed{{\lambda }_{\pm }=u\pm 2\sqrt{v}}$, with left eigenvectors ${\ell }_{\pm }^T=(1,\pm 1/\sqrt{v})$. Thus the system is strictly hyperbolic there; at $v=0$ the eigenvalues coincide and the matrix is defective, so strict hyperbolicity fails. The characteristic directions are $dy/dx={\lambda }_{\pm }$.

If $\dot{y}={\lambda }_{\pm }\dot{x}$, multiplication of $\dot{U}=U_x\dot{x}+U_y\dot{y}$ by the left eigenvector gives ${\ell }_{\pm }^T\dot{U}={\ell }_{\pm }^{T}C\dot{x}$. Since $d(u\pm 2\sqrt{v})=du\pm dv/\sqrt{v}$, defineThenConsequently $R$ is constant on the minus characteristic curves, and $S-2x$ is constant on the plus characteristic curves. One reconstructs $\boxed{u=(R+S)/2,v=(S-R{)}^2/16}$, with the essential physical constraint $S\ge R$.

For the data on $y=0$, the nonnegative square root givesThus, on any regular patch, use the two footpoints determined byand put $R=R_0(\xi )$, $S=2(x-\eta )+S_0(\eta )$, selecting branches continuous from the initial point. These implicit algebraic formulas determine the local answer wherever both characteristic curve projections are regular and $S>R$.

For example, near an initial point with $x>1$, they give $R=3/2$, $S=2x-1/2$, so $\boxed{u=x+1/2,v=(x-1{)}^2/4}$. On an initial interval $x<1$, define $H=\sqrt{9/4+4y}$. One instead has $S=H$ andnear the chosen initial point $x_0\ne 1/4$. The radicands must be positive, the footpoints must stay in their indicated initial intervals, and $H>R$ must hold.

There is no continuously differentiable solution realizing the first data set on the entire stated initial line. At $(x,y)=(1/4,0)$, $u=3/4$, $\sqrt{v}=3/8$, $u_x=1$ and $v_x=-3/8$. The minus left eigenvector is $(1,-8/3)$ and its eigenvalue is zero. The differential equation there would require ${\ell }_{-}^T{U}_x={\ell }_{-}^{T}C$, but its two sides are respectively $2$ and $0$. This is an explicit compatibility contradiction, not merely a loss of a preferred formula. Also, the proposed simple expression independent of $y$ does not solve the equation for $x<1$, because then $v_x<0$ but $\sqrt{v}>0$.

For the second data set, treat $x$ as the evolution variable and let $\eta$ label the initial $y$ coordinate. Then $R_0(\eta )=4\min (\eta ,0)$ and $S_0(\eta )=4\max (\eta ,0)$. In the positive quadrant the minus characteristic curves give $R=0$. The plus characteristic curves satisfy $S=2x+S_0(\eta )$ and $y=\eta +x^2+x{S}_0(\eta )$. Therefore their continuous characteristic curve continuation isAbove the parabola the footpoint is $\eta =(y-x^2)/(1+4x)>0$; below it the plus footpoint is $\eta =y-x^2<0$. Each branch directly satisfies both equations and $v>0$. They agree continuously on $y=x^2$, but their $y$ derivatives are $0$ and $2/(1+4x)$, respectively. Thus the continuation is piecewise classical and not continuously differentiable across the parabola. The source's globally continuously differentiable formulation needs this qualification. If only positive initial $y$ is intended, the upper rational formula is a classical continuation throughout the positive quadrant, but the region $y<x^2$ requires incoming information from $y=0$ and is not determined by that positive half of the initial line alone. This illustrates loss of differentiability from square-root hyperbolic source data.

A first-order system $U_t+A(U)U_x=C(U)$ is strictly hyperbolic at a state when $A$ has distinct real eigenvalues. These separate its characteristic families and permit independent compatibility equations from left eigenvectors. At a repeated eigenvalue strict hyperbolicity fails; if the matrix is also defective, even hyperbolicity in the complete-eigenvector sense fails. For $A=\left(\begin{array}{cc}u& 2\\ 2v& u\end{array}\right)$ the eigenvalues $u\pm 2\sqrt{v}$ are distinct precisely when $v>0$.