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# Download An Introduction to Partial Differential Equations by Y. Pinchover, J. Rubenstein PDF

By Y. Pinchover, J. Rubenstein

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Example text

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. 29) Thus we expect a unique solution (at least locally) near each point on the initial curve, except, perhaps, the point x = 0.

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. 16). Assume further that the transversality condition holds at each point s in the interval (s0 − 2δ, s0 + 2δ) on the initial curve.

4). Therefore, each one of them intersects the projection of the initial curve (the x axis) twice. We also saw that the Jacobian vanishes at the origin. So how is it that we seem to have obtained a unique solution? The mystery is easily resolved by observing that in choosing the positive sign for the square root in the argument of ψ, we effectively reduced the solution to the ray {x > 0}. Indeed, in this region a characteristic intersects the projection of the initial curve only once. 7 Solve the equation u x + 3y 2/3 u y = 2 subject to the initial condition u(x, 1) = 1 + x.