IVP of quasi-linear transport equation

Question

I need help second part of this problem: Show that the solution $u$ of the quasi-linear PDE:

$u_y + a(u)u_x=0$

with initial condition $u(x,0)=h(x)$ is given implicitly by

$u=h(x-a(u)y)$

Show that the solution becomes singular for some positive $y$, unless $a(h(s))$ is a nondecreasing function of $s$

Here what I got the first part:

$u_y + a(u)u_x=0$

$u(x,0)=h(x)$

The initial curve $\Gamma : <x=s, y=0, z=h(s)>$ and the characteristic differential equations are $\frac{dx}{dt}=a(u), \frac{dy}{dt}=1, \frac{dz}{dt}=0$ .This leads to the parametric representation:

$x=a(u) + t$, $y=t$, $z=h(s)$. We solve and get $ s=x-a(u)y$ and solution of the initial value problem above is $z=h(s)=h(x-a(u)y)$.

For second part, I am not sure what I did: The characteristic projection $C_s$ in the $xy$-plane passing through the point $(s,0)$ is the line

$ x= s + a(u)y $ along which $u$ has the constant value $u=h(s)$ . Now two characteristics $C_{s_1}$ and $C_{s_2}$ intersect at a point $(x,y)$ with : $y=-\frac{s_2 - s_1}{a(h(s_2))-a(h(s_1))}$ (1). If $s_2\neq s_1$ and $a(h(s_2))\neq a(h(s_1))$, the function $u$ must take the distinct values $a(h(s_1))$ and $a(h(s_2))$ at $(x,y)$ and hence cannot be univalued. There always exist positive y of (1), unless $a(h(s))$ is a nondecreasing function of s. For all other $a(h(s))$ the solution $u(x,y)$ becomes singular for some positive y.

Please advise. Thanks

Answer 1

The proof is fine, since the statement "the slope of the chords of $a\circ h$ is non-negative" implies "$a\circ h$ is non-decreasing".

Now, what happens if $a\circ h$ is decreasing somewhere? Using the mean value theorem over $[s_1,s_2]$, we rewrite the intersection $(1)$ in OP as $$ y = -\frac{s_2-s_1}{a\circ h(s_2)-a\circ h(s_1)} = -\frac{1}{(a\circ h)'(s)} $$ for some $s \in [s_1,s_2]$. Now, we obtain that the smallest $y>0$ at which characteristics intersect is given by $$ y_b = \inf_{s\in\Bbb R} \left( -\frac{1}{(a\circ h)'(s)}\right) = \frac{-1}{\inf_{s\in \Bbb R} (a\circ h)'(s)}\, . $$ This is the $y$ where a shock wave occurs (breaking time).

There are several other proofs than the previous geometric argument (cf. e.g. (1)), which is based on the intersection of characteristics and the mean value theorem. Let us summarize a few methods:

1. Dependence to initial data. We differentiate the expression of characteristics $x = x_0 + a\circ h(x_0)\, y$ with respect to the initial abscissa $x_0$. This derivative vanishes at $y = -1/(a\circ h)'(x_0)$. The smallest such $y$ is the breaking time.
2. Characteristic evolution of $u_x$. We differentiate the PDE with respect to $x$, and we set $q=u_x$, $q' = q_y + a(u) q_x$. Hence, $q' + a'(u)\, q^2 = 0$, which solution is $q(y) = q_0/(1+q_0 k y)$, where $q_0 = h'(x_0)$ and $k = a'(u) = a'\circ h(x_0)$ is constant along the characteristics. Finally, $q$ blows up at $y = -1/(q_0 k) = -1/(a\circ h)'(x_0)$.
3. Implicit function theorem. The implicit equation $u = h(x-a(u) y)$ can be solved for $u$ as a differentiable function of $x$ and $y$ for $y$ small enough. Differentiating w.r.t. $x$, one has $u_x = q_0/(1+q_0 k y)$ with the same notations as above.

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(1) P.D. Lax, Hyperbolic systems of conservation laws and the mathematical theory of shock waves. SIAM, 1973. doi:10.1137/1.9781611970562.ch1