By Yehuda Pinchover and Jacob Rubinstein

ISBN-10: 0511111576

ISBN-13: 9780511111570

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**Example text**

We obtain y = (x + 1 − s)3 , and, thus, for each ﬁxed s this is an equation for a characteristic. 5. While the picture indicates no problems, we were not careful enough in solving the characteristic equations, since the function y 2/3 is not Lipschitz continuous at the origin. Thus the characteristic equations might not have a unique solution there! In fact, it can be easily veriﬁed that y = 0 is also a solution of yt = 3y 2/3 . 5, the well behaved characteristics near the projection of the initial curve y = 1 intersect at some point the extra characteristic y = 0.

B) 3u x − 7u y = 0. (c) 2u x + π u y = 0. 2 Show that each of the following equations has a solution of the form u(x, y) = eαx+βy . Find the constants α, β for each example. (a) u x + 3u y + u = 0. (b) u x x + u yy = 5ex−2y . (c) u x x x x + u yyyy + 2u x x yy = 0. 54) together with the initial condition u(0, 0) = 0. 55) has no solution at all. 4 Let u(x, y) = h( x 2 + y 2 ) be a solution of the minimal surface equation. (a) Show that h(r ) satisﬁes the ODE r h + h (1 + (h )2 ) = 0. (b) What is the general solution to the equation of part (a)?

Namely, the transversality condition does not hold there (a fact we already expected from our computation of the Jacobian above). Indeed the violation of the transversality condition led to nonuniqueness of the solution near the curve {(x, y) | x < 0 and y = x + x 2 }, which is manifested in the ambiguity of the sign of the square root. 6 Solve the equation −yu x + xu y = u subject to the initial condition u(x, 0) = ψ(x). 27) x(0, s) = s, y(0, s) = 0, u(0, s) = ψ(s). 28) Let us examine the transversality condition: J= 0 s = −s.

### An introduction to partial differential equations by Yehuda Pinchover and Jacob Rubinstein

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