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By Y. Pinchover, J. Rubenstein

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Returning to (i) we obtain x(t, s) = (1 + s)et − e−t and u(t, s) = set + e−t . Observing that x − y = set − e−t , we finally get u = 2/y + (x − y). The solution is not global (it becomes singular on the x axis), but it is well defined near the initial curve. 5 The existence and uniqueness theorem We shall summarize the discussion on linear and quasilinear equations into a general theorem. For this purpose we need the following definition. 16) defining an initial curve for the integral surface. e. J |t=0 = xt (0, s)ys (0, s) − yt (0, s)xs (0, s) = a b = 0.

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.

5 Solve the equation u x + u y + u = 1, subject to the initial condition u = sin x, on y = x + x 2 , x > 0. 25) 2 respectively. Let us compute first the Jacobian along the initial curve: J= 1 1 = 2s. 26) Thus we anticipate a unique solution at each point where s = 0. Since we are limited to the regime x > 0 we indeed expect a unique solution. The parametric integral surface is given by (x(t, s), y(t, s), u(t, s)) = (s + t, s + s 2 + t, 1 − (1 − sin s)e−t ). In order to invert the mapping (x(t, s), y(t, s)), we substitute the equation for x into the equation for y to obtain s = (y − x)1/2 .

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