By Y. Pinchover, J. Rubenstein

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**Sample text**

We also notice that, in general, the parameterization (x(t, s), y(t, s), u(t, s)) represents a surface in R3 . 3) as well. Namely, each point on the initial curve is a starting point for a characteristic curve. 15) supplemented by the initial condition x(0, s) = x0 (s), y(0, s) = y0 (s), u(0, s) = u 0 (s). 16) is called the Cauchy problem for quasilinear equations. 13) are independent of the third equation and of the initial conditions. We shall observe later the special role played by the projection of the characteristic curves on the (x, y) plane.

7 Solve the equation u x + 3y 2/3 u y = 2 subject to the initial condition u(x, 1) = 1 + x. 30) x(0, s) = s, y(0, s) = 1, u(0, s) = 1 + s. 31) In this example we expect a unique solution in a neighborhood of the initial curve since the transversality condition holds: J= 1 3 = −3 = 0. 32) The parametric integral surface is given by x(t, s) = s + t, y(t, s) = (t + 1)3 , u(t, s) = 2t + 1 + s. Before proceeding to compute an explicit solution, let us ﬁnd the characteristics. For this purpose recall that each characteristic curve passes through a speciﬁc s value.

We conclude that the solution’s derivative blows up at the critical time yc = − 1 . 47) Hence the classical solution is not deﬁned for y > yc . This conclusion is consistent with the heuristic physical interpretation presented above. Indeed a necessary condition for a singularity formation is that h (s) < 0 at least at one point, such that a faster characteristic will start from a point behind a slower characteristic. If h(s) is never decreasing, there will be no singularity; however, such data are exceptional.