By Y. Pinchover, J. Rubenstein

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**Additional resources for An Introduction to Partial Differential Equations**

**Example text**

7) we obtain that u is constant along the lines 32 y + (−x)3/2 = constant. 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 ).

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. 4 Examples of the characteristics method 35 y char. projection of Γ 1 x char. char. 5 Self-intersection of characteristics. for y. We obtain y = (x + 1 − s)3 , and, thus, for each ﬁxed s this is an equation for a characteristic.

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.